Almost half of numbers are divisible by two.

For the string shouldn't be any i ≤ n - 2, s_i = s_i + 2. Make some examples of little sizes.

Find minimum weighted edges.

There is some edges with weight 0, try to connect them in a carefully way.

The last situation is some 'a' characters after some 'b' ones.

The last situation is unique.

The number of steps is also unique.

Each 'b' character makes a number of 'b' characters in the last situation according to the number of 'a' characters before it.

We want to color the ice creams in a way that all of ice creams in a vertex would be different with this condition that if two vertices u and v have ice cream i, then all of vertices in the unique path of u and v would have it.

Let's have a trivial lower-bound for the answer.

You can prove that by induction on leaves of the tree. And code it how you proved it.

Let's solve the problem for just two tree.

consider this trivial solution and try to improve it to .

You can prove, we need even number of swaps to reach the same permutation. So and the answer for 4·k + 3 and 4·k + 4 is -1.

The solution almost is constructive.

Let's partition n = 4·k numbers to classes, each class contains a 4 consecutive numbers. And when n = 4·k + 1 there is a class with 5 consecutive numbers.

The problem consists two parts, the first part is to find mn_i and mx_i, the minimum and the maximum score i-th vertex may get after the score distribution. The second part is to finding the answer and all your need to solve this part is those two arrays. It is trivial mn_i is number of real bullion of golds in i-th vertex.

Let's find the scores for two vertices u and v, u have a directed edge to v.

u -> v

If i-th element from u-th vertex was painted and j-th element from v-th vertex was unpainted such that then j-th element from v-th vertex will be painted quickly. For two vertex, we can do it in complexity .

If we have this directed walk {w₁, w₂, ..., w_k}, you should affect v₁ on v_k how I said in the previous hint with .

What about strogly connected components.

In this components if i-th element from u-th vertex was painted then for all of j such that , j-th element from all vertices will be painted quickly. For this components, we can do it with complexity . In the main problem, we can suppose all of a strongly conected component a vertex with size g in this way and we can calculate mx_i with answer of its component's vertex multiplied by .

SCC of the tournament yields a Hamiltonian path of strongly connected components( if we consider each component a vertex ). The problem decreased to solve this path. we can solve the path with the complexity of sum of vertices' sizes.

Now, for every vertex we have minimum and maximum score it may get. For B known vertices, to check if they can be the wanted set, we can assume their score be maximum possible and the other scores be minimum possible. If they were top, then they can be a sample of the wanted set.

Consider a sorted sequence of combination of two array mn and mx.

For every vertex, consider the number of ways it(with its maximum score) can be the minimum score of the set of B vertices.

l,r=list(map(int,input().split()))
print(2 if l<r else l)

int main(){
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d", l == r ? l : 2);
}

print(("aabb" * 50000)[:int(input())])

int N;
int main()
{
	scanf("%d", &N);
	for (int i = 0; i < N; i++)
		putchar(i & 2 ? 'b' : 'a');
	puts("");
}

print((int(input())-1)//2)

int main(){
	int n;
	scanf("%d", &n);
	printf("%d\n", (n-1)/2);
}

s = input()
cnt = 0
m=10**9 + 7
t = 0

for i in range(len(s)):
	if s[~i] == 'a':
		cnt = (cnt+t)%m
		t = (t*2)%m
	else:
		t += 1
print(cnt)

int c,d,i,n,m,k,x,j=1000000007;
string s;
main(){
	cin>>s;
	for(i=s.size()-1;i>=0;i--){
		if(s[i]=='b')c++;else{
			k+=c;c*=2;k%=j;c%=j;
		}
	}
	cout<<k;
}

//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

int const N = 3e5 + 10 ;
int n , m , ans = 1 , res[N] ;
vector <int> g[N] , vec[N] ;

inline bool cmp (int a , int b) {
	return res[a] < res[b] ;
}

void dfs (int v , int par = -1) {
	int col = ans  , p = (int)vec[v].size() - 1 ;

	sort(vec[v].begin() , vec[v].end() , cmp) ;

	for (int x : vec[v]) {
		if (res[x]) {
			continue ;
		}

		while (p >= 0 && col == res[vec[v][p]]) {			
			col -- ;
			p -- ;
		}

		res[x] = col ;
		col -- ;
	
	}
	
	for (int u : g[v]) {
		if (u != par) {
			dfs(u , v) ;
		}
	}
}

int main(){
	ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) ;

	cin >> n >> m ;
	
	for (int i = 0 ; i < n ; i ++) {
		int sz ;
		cin >> sz ;

		ans = max(ans , sz) ;

		while (sz --) {
			int x ;
			cin >> x ;
			x -- ;
			vec[i].push_back(x) ;
		}
	}

	for (int i = 0 ; i < n - 1 ; i ++) {
		int u , v ;
		cin >> u >> v ;
		u -- , v -- ;

		g[u].push_back(v) ;
		g[v].push_back(u) ;
	}

	dfs(0) ;

	cout << ans << '\n' ;

	for (int i = 0 ; i < m ; i ++) {
		cout << max(res[i] , 1) << ' ' ;
	}
	cout << '\n' ;
}

vector <int> Vs[300050];
vector <int> conn[300050];

int ans[300050];
bool chk[500050];
bool dchk[300050];
vector <int> Vu;
void DFS(int n) {
	dchk[n] = true;

	for (auto it : Vs[n]) {
		if (ans[it]) chk[ans[it]] = true;
		else Vu.push_back(it);
	}

	int a = 1;
	for (auto it : Vu) {
		while (chk[a]) a++;
		ans[it] = a++;
	}
	for (auto it : Vs[n]) chk[ans[it]] = false;
	Vu.clear();

	for (auto it : conn[n]) {
		if (dchk[it]) continue;
		DFS(it);
	}
}
int main() {
	int N, M, i;
	scanf("%d %d", &N, &M);
	for (i = 1; i <= N; i++) {
		int t1, t2;
		scanf("%d", &t1);
		while (t1--) {
			scanf("%d", &t2);
			Vs[i].push_back(t2);
		}
	}
	for (i = 1; i < N; i++) {
		int t1, t2;
		scanf("%d %d", &t1, &t2);
		conn[t1].push_back(t2);
		conn[t2].push_back(t1);
	}
	DFS(1);

	int mx = 0;
	for (i = 1; i <= M; i++) {
		mx = max(mx, ans[i]);
		if (ans[i] == 0) ans[i] = 1;
	}
	mx = max(mx, 1);
	printf("%d\n", mx);
	for (i = 1; i <= M; i++) printf("%d ", ans[i]);
	
	return !printf("\n");
}

using namespace std ;

#define int long long

const int MAXN = 1e5 + 100 ;

vector<int>ver[MAXN] , com[MAXN] , ps[MAXN] ; 
int vis[MAXN] , dis[MAXN] ; 
int mxr , mxid ; 
void dfs(int v , int col = -1 , int h = 0 , int par = -1){
	vis[v] = max(vis[v] , col) ; 
	if(mxr < h)mxid = v , mxr = h ; 
	dis[v] = max(h , dis[v]) ; 
	if(col > 0)com[col] . push_back(dis[v]) ; 
	for(auto u : ver[v]){
		if(u == par)continue ; 
		dfs(u , col , h + 1 , v) ; 
	}
}
map<pair<int , int> , int> check ;
int32_t main(){
    ios_base::sync_with_stdio(0) ;
    cin . tie(0) ; cout . tie(0) ;
    int n , m , q ; cin >> n >> m >> q ; 
    for(int i = 0 ; i < m ; i ++){
    	int x , y ; cin >> x >> y ; 
    	x -- , y -- ; 
    	ver[x] . push_back(y) ; 
    	ver[y] . push_back(x) ;
    }
    int cnt = 0 ; 
    for(int i = 0 ; i < n ; i ++){
    	if(vis[i])continue ; 
    	mxr = 0 , mxid = i ; 
    	dfs(i) ; 
    	mxr = 0 ; 
    	dfs(mxid) ;
    	dfs(mxid , ++ cnt) ; sort(com[cnt] . begin() , com[cnt] . end()) ; 
    	ps[cnt] . push_back(0) ; 
    	for(auto v : com[cnt])ps[cnt] . push_back(ps[cnt] . back() + v) ; 
    }
	cout << fixed << setprecision(10) ;
    while(q --){
    	int x , y ; cin >> x >> y ; 
    	x -- , y -- ; x = vis[x] ; y = vis[y] ; 
    	if(com[x] . size() > com[y] . size())swap(x , y) ; 
    	if(x == y){cout << -1 << '\n' ; continue ; }
    	if(check[{x , y}] > 0){cout << double(check[{x , y}]) / (1ll * com[x] . size() * com[y] . size()) << '\n' ; continue ; }
    	int ans = 0 , mxd = max(com[x] . back() , com[y] . back()) ; 
    	for(auto v : com[x]){
    		int num = lower_bound(com[y] . begin() , com[y] . end() , mxd - v - 1) - com[y] . begin() ; 
    		ans += num * mxd + (com[y] . size() - num) *  (v + 1) + ps[y] . back() - ps[y][num] ; 
    	}
    	check[{x , y}] = ans ; 
    	cout << double(check[{x , y}]) / (1ll * com[x] . size() * com[y] . size()) << '\n' ; 
    }
}

int n;

int main()
{
	scanf("%d",&n);
	if(n%4>1)return printf("NO\n"),0;
	printf("YES\n");
	for(int i=1;i<n;i+=4)
	{
		if(n%4)printf("%d %d\n%d %d\n%d %d\n",i+2,n,i+2,i+3,i+3,n);
		else printf("%d %d\n",i+2,i+3);
		printf("%d %d\n",i,i+2);
		printf("%d %d\n",i+1,i+3);
		printf("%d %d\n",i+1,i+2);
		printf("%d %d\n",i,i+3);
		if(n%4)printf("%d %d\n%d %d\n%d %d\n",i,n,i,i+1,i+1,n);
		else printf("%d %d\n",i,i+1);
	}
	for(int i=1;i<n;i+=4)
		for(int j=i+4;j<n;j+=4)
		{
			printf("%d %d\n",i+3,j+2);
			printf("%d %d\n",i+2,j+2);
			printf("%d %d\n",i+3,j+1);
			printf("%d %d\n",i,j+1);
			printf("%d %d\n",i+1,j+3);
			printf("%d %d\n",i+2,j+3);
			printf("%d %d\n",i+1,j+2);
			printf("%d %d\n",i+1,j+1);
			printf("%d %d\n",i+3,j);
			printf("%d %d\n",i+3,j+3);
			printf("%d %d\n",i,j+2);
			printf("%d %d\n",i,j+3);
			printf("%d %d\n",i+2,j);
			printf("%d %d\n",i+2,j+1);
			printf("%d %d\n",i+1,j);
			printf("%d %d\n",i,j);
		}
	return 0;
}

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.Stack;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.Stream;
import java.util.Vector;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.BitSet;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        TaskF solver = new TaskF();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskF {
        public int mod = 1000000007;
        public int n;
        public int a;
        public int b;
        public char[][] adj;
        public List<Integer>[] graph;

        public void solve(int testNumber, InputReader in, OutputWriter out) {
            n = in.nextInt();
            a = in.nextInt();
            b = in.nextInt();
            adj = new char[n][n];
            graph = Stream.generate(ArrayList::new).limit(n).toArray(List[]::new);
            for (int i = 0; i < n; i++) adj[i] = in.next().toCharArray();
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (adj[i][j] == '1') {
                        graph[i].add(j);
                    }
                }
            }
            List<List<Integer>> comp = new SCCTarjan().scc(graph);
            boolean[][] can = new boolean[n][];
            int[] low = new int[n];
            for (int i = 0; i < n; i++) {
                int q = in.nextInt();
                char[] w = in.next().toCharArray();
                can[i] = new boolean[q];
                for (int j = 0; j < q; j++) {
                    can[i][j] = w[j] == '1';
                    if (can[i][j]) low[i]++;
                }
            }
            int[] wcomp = new int[n];
            int idx = 0;
            List<BitSet> bb = new ArrayList<>();
            List<Integer> bs = new ArrayList<>();
            for (List<Integer> gg : comp) {
                int gcd = 0;
                for (int x : gg) {
                    gcd = Utils.gcd(gcd, can[x].length);
                    wcomp[x] = idx;
                }
                idx++;
                BitSet d = new BitSet(gcd);
                bs.add(gcd);
                bb.add(d);
                for (int x : gg) {
                    for (int i = 0; i < can[x].length; i++) {
                        if (can[x][i]) {
                            d.set(i % gcd, true);
                        }
                    }
                }
            }

            for (int kk = comp.size() - 1; kk > 0; kk--) {
                int g = Utils.gcd(bs.get(kk), bs.get(kk - 1));
                BitSet ww = new BitSet(g);
                for (int i = 0; i < bs.get(kk); i += g) {
                    ww.or(bb.get(kk).get(i, i + g));
                }
                for (int i = 0; i < bs.get(kk - 1); i++) {
                    if (ww.get(i % g)) {
                        bb.get(kk - 1).set(i, true);
                    }
                }
            }
            int[] high = new int[n];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < can[i].length; j++) {
                    if (bb.get(wcomp[i]).get(j % bs.get(wcomp[i])))
                        high[i]++;
                }
            }
            long ret = 0;
            int[][] comb = Utils.getComb(n + 1, mod);
            for (int lowest = 0; lowest < n; lowest++) {
                int countabove = 0;
                for (int other = 0; other < n; other++) {
                    if (other == lowest) continue;
                    if (low[other] > high[lowest]) {
                        countabove++;
                    }
                }
                if (countabove >= a) {
                    continue;
                }
                int canbig = 0;
                for (int other = 0; other < n; other++) {
                    if (other == lowest) continue;
                    if (low[other] <= high[lowest] && (high[other] > high[lowest] || (high[other] == high[lowest] && other > lowest))) {
                        canbig++;
                    }
                }

                for (int take = 0; take < b && take <= canbig && take + countabove < a; take++) {
                    ret = (ret + 1L * comb[canbig][take] * comb[countabove][b - take - 1]) % mod;
                }
            }
            out.println(ret);
        }

    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void close() {
            writer.close();
        }

        public void println(long i) {
            writer.println(i);
        }

    }

    static class Utils {
        public static int gcd(int a, int b) {
            return b == 0 ? a : gcd(b, a % b);
        }

        public static int[][] getComb(int sz, int mod) {
            int[][] comb = new int[sz][sz];
            for (int i = 0; i < sz; i++) {
                comb[i][0] = 1;
                for (int j = 1; j <= i; j++) {
                    comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
                    if (comb[i][j] >= mod) comb[i][j] -= mod;
                }
            }
            return comb;
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (this.numChars == -1) {
                throw new InputMismatchException();
            } else {
                if (this.curChar >= this.numChars) {
                    this.curChar = 0;

                    try {
                        this.numChars = this.stream.read(this.buf);
                    } catch (IOException var2) {
                        throw new InputMismatchException();
                    }

                    if (this.numChars <= 0) {
                        return -1;
                    }
                }

                return this.buf[this.curChar++];
            }
        }

        public int nextInt() {
            int c;
            for (c = this.read(); isSpaceChar(c); c = this.read()) {
                ;
            }

            byte sgn = 1;
            if (c == 45) {
                sgn = -1;
                c = this.read();
            }

            int res = 0;

            while (c >= 48 && c <= 57) {
                res *= 10;
                res += c - 48;
                c = this.read();
                if (isSpaceChar(c)) {
                    return res * sgn;
                }
            }

            throw new InputMismatchException();
        }

        public String next() {
            int c;
            while (isSpaceChar(c = this.read())) {
                ;
            }

            StringBuilder result = new StringBuilder();
            result.appendCodePoint(c);

            while (!isSpaceChar(c = this.read())) {
                result.appendCodePoint(c);
            }

            return result.toString();
        }

        public static boolean isSpaceChar(int c) {
            return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
        }

    }

    static class SCCTarjan {
        List<Integer>[] graph;
        boolean[] visited;
        Stack<Integer> stack;
        int time;
        int[] lowlink;
        List<List<Integer>> components;

        public List<List<Integer>> scc(List<Integer>[] graph) {
            int n = graph.length;
            this.graph = graph;
            visited = new boolean[n];
            stack = new Stack<>();
            time = 0;
            lowlink = new int[n];
            components = new ArrayList<>();

            for (int u = 0; u < n; u++)
                if (!visited[u])
                    dfs(u);

            return components;
        }

        void dfs(int u) {
            lowlink[u] = time++;
            visited[u] = true;
            stack.add(u);
            boolean isComponentRoot = true;

            for (int v : graph[u]) {
                if (!visited[v])
                    dfs(v);
                if (lowlink[u] > lowlink[v]) {
                    lowlink[u] = lowlink[v];
                    isComponentRoot = false;
                }
            }

            if (isComponentRoot) {
                List<Integer> component = new ArrayList<>();
                while (true) {
                    int x = stack.pop();
                    component.add(x);
                    lowlink[x] = Integer.MAX_VALUE;
                    if (x == u)
                        break;
                }
                components.add(component);
            }
        }

    }
}

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

const int maxn = 5005;
const int maxelem = 2000006;
const int MOD = 1000000007;

char gr[maxn][maxn];
string init[maxn];
int mincnt[maxn], maxcnt[maxn];
int c[maxn][maxn];
int n, m, A, B;
bool was[maxn];
vector<int> order;
vector<bool> can[maxn];
bool nowcan[maxelem];
char s[maxelem];
int gcd[maxn];
vector<int> comp[maxn];

void preorder(int cur)
{
    if (was[cur]) return;
    was[cur] = true;
    for (int i = 0; i < n; i++) if (gr[cur][i] == '1') preorder(i);
    order.pb(cur);
}

void color(int cur, int cc)
{
    if (was[cur]) return;
    was[cur] = true;
    gcd[cc] = __gcd(gcd[cc], (int)init[cur].size());
    comp[cc].pb(cur);
    for (int i = 0; i < n; i++) if (gr[i][cur] == '1') color(i, cc);
}

int getc(int n, int k)
{
    if (k < 0 || k > n) return 0;
    return c[n][k];
}

int main()
{
    scanf("%d%d%d", &n, &A, &B);
    for (int i = 0; i < n; i++)
    {
        scanf("%s", gr[i]);
    }
    for (int i = 0; i < n; i++)
    {
        scanf("%*d");
        scanf("%s", s);
        init[i] = string(s);
        for (int j = 0; j < (int)init[i].size(); j++) mincnt[i] += init[i][j] == '1';
        maxcnt[i] = mincnt[i];
    }
    for (int i = 0; i < n; i++) if (!was[i]) preorder(i);
    memset(was, 0, sizeof(was));
    int cc = 0;
    reverse(all(order));
    for (auto t : order) if (!was[t])
    {
        color(t, cc);
        cc++;
    }
    for (int i = 0; i < cc; i++)
    {
        can[i].resize(gcd[i]);
        for (int j = 0; j < gcd[i]; j++) can[i][j] = false;
        for (auto t : comp[i])
        {
            for (int j = 0; j < (int)init[t].size(); j++) if (init[t][j] == '1') can[i][j % gcd[i]] = true;
        }
    }
    for (int i = 0; i < cc - 1; i++)
    {
        int g = __gcd(gcd[i], gcd[i + 1]);
        for (int j = 0; j < g; j++) nowcan[j] = false;
        for (int j = 0; j < gcd[i]; j++) nowcan[j % g] |= can[i][j];
        for (int j = 0; j < gcd[i + 1]; j++) can[i  + 1][j] = (can[i  + 1][j] | nowcan[j % g]);
    }
    for (int i = 0; i < cc; i++)
    {
        for (auto t : comp[i])
        {
            for (int j = 0; j < (int)init[t].size(); j++) if (init[t][j] == '0' && can[i][j % gcd[i]]) maxcnt[t]++;
        }
    }
    
    c[0][0] = 1;
    for (int i = 1; i <= n; i++)
    {
        c[i][0] = 1;
        for (int j = 1; j <= i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
    }
    int answer = 0;
    for (int i = 0; i < n; i++)
    {
        int cntbigger = 0;
        int cntupper = 0;
        for (int j = 0; j < n; j++) if (i != j)
        {
            if (mincnt[j] > maxcnt[i]) cntbigger++;
            else if (maxcnt[j] > maxcnt[i] || (maxcnt[j] == maxcnt[i] && j > i)) cntupper++;
        }
        if (cntbigger >= A) continue;
        for (int j = max(0, B - 1 - cntbigger); j <= cntupper && j + cntbigger + 1 <= A && j + 1 <= B; j++)
        {
            answer = (answer + (ll)getc(cntupper, j) * getc(cntbigger, B - 1 - j)) % MOD;
        }
    }
    cout << answer << endl;
    return 0;
}	 

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define fill( x, y ) memset( x, y, sizeof x )
#define copy( x, y ) memcpy( x, y, sizeof x )
using namespace std;

typedef long long LL;
typedef pair < int, int > pa;

inline int read()
{
	int sc = 0, f = 1; char ch = getchar();
	while( ch < '0' || ch > '9' ) { if( ch == '-' ) f = -1; ch = getchar(); }
	while( ch >= '0' && ch <= '9' ) sc = sc * 10 + ch - '0', ch = getchar();
	return sc * f;
}

const int MAXN = 5005;
const int MAXM = 2000005;
const int mod = 1e9 + 7;

int n, a, b, len[MAXN], can[MAXM], id_cnt, L[MAXN], R[MAXN], C[MAXN][MAXN];
int dfn[MAXN], scc[MAXN], low[MAXN], tim, num, st[MAXN], top, w[MAXN], ans;
vector < int > G[MAXN], id[MAXN], scc_bit[MAXN], v[MAXN];
char ch[MAXM];

inline void inc(int &x, int y) { x += y; if( x >= mod ) x -= mod; }

inline void dfs(int x)
{
	dfn[ x ] = low[ x ] = ++tim; st[ ++top ] = x;
	for( auto y : G[ x ] )
		if( !dfn[ y ] ) dfs( y ), low[ x ] = min( low[ x ], low[ y ] );
		else if( !scc[ y ] ) low[ x ] = min( low[ x ], dfn[ y ] );
	if( dfn[ x ] == low[ x ] )
	{
		num++; int t = 0;
		while( t ^ x )
		{
			scc[ t = st[ top-- ] ] = num;
			w[ num ] = __gcd( w[ num ], len[ t ] );
			v[ num ].pb( t );
		}
		scc_bit[ num ].resize( w[ num ] );
		for( auto y : v[ num ] )
			for( int d = 0 ; d < len[ y ] ; d++ ) if( can[ id[ y ][ d ] ] ) scc_bit[ num ][ d % w[ num ] ] = 1;
	}
}

int main()
{
#ifdef wxh010910
	freopen( "data.in", "r", stdin );
#endif
	n = read(), a = read(), b = read();
	for( int i = 0 ; i <= n ; i++ )
	{
		C[ i ][ 0 ] = 1;
		for( int j = 1 ; j <= i ; j++ ) inc( C[ i ][ j ] = C[ i - 1 ][ j ], C[ i - 1 ][ j - 1 ] );
	}
	for( int i = 1 ; i <= n ; i++ )
	{
		scanf( "%s", ch + 1 );
		for( int j = 1 ; j <= n ; j++ ) if( ch[ j ] == '1' ) G[ i ].pb( j );
	}
	for( int i = 1 ; i <= n ; i++ )
	{
		len[ i ] = read(); scanf( "%s", ch );
		for( int j = 0 ; j < len[ i ] ; j++ ) id[ i ].pb( ++id_cnt ), can[ id_cnt ] = ch[ j ] == '1', L[ i ] += can[ id_cnt ];
	}
	for( int i = 1 ; i <= n ; i++ ) if( !dfn[ i ] ) dfs( i );
	for( int i = num ; i > 1 ; i-- )
	{
		int g = __gcd( w[ i ], w[ i - 1 ] );
		w[ i - 1 ] = g;
		for( int j = 0 ; j < w[ i ] ; j++ )	scc_bit[ i - 1 ][ j % g ] |= scc_bit[ i ][ j ];
	}
	for( int i = 1 ; i <= n ; i++ )
	{
		int x = scc[ i ];
		for( int j = 0 ; j < len[ i ] ; j++ )
			R[ i ] += scc_bit[ x ][ j % w[ x ] ];
	}
	for( int i = 1 ; i <= n ; i++ )
	{
		int t1 = 0, t2 = 0;
		for( int j = 1 ; j <= n ; j++ ) if( L[ j ] > R[ i ] ) t1++;
		if( t1 >= a ) continue;
		for( int j = 1 ; j <= n ; j++ ) if( L[ j ] <= R[ i ] && mp( R[ j ], j ) > mp( R[ i ], i ) ) t2++;
		for( int j = 0 ; j < b && j <= t2 && j + t1 < a ; j++ )
		   inc( ans, 1LL * C[ t2 ][ j ] * C[ t1 ][ b - j - 1 ] % mod );	
	}
	return printf( "%d\n", ans ), 0;
}

Consider n - f_n , it is the minimum period, let's call it A. It makes a good look on the problem.

You may have a sense of recursive solving the problem as you see we almost minimized the problem in the minimum period. I agree with you, but see more carefully.

//    whoisHussain.org
#include <bits/stdc++.h>
using namespace std;

const int N = 300011;
int f[N],q, g[N],sz; 
bool a[N];
int32_t main(){
        string s;
        int n;
        ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
        cin >> q;
        while( q-- ){
                cin >> s;
                n = s.size(), sz = 0;

                f[0] = f[1] = 0;
                for(int i = 2,j = 0; i <= n; i++){
                        while( j!=0 && s[i-1] != s[j] )
                                j = f[j]; 
                        if( s[i-1]==s[j] ) j++;
                        f[i] = j;
                }

                for(int i = n; i; i = f[i] ) g[sz++] = n-i;
                
                for(int i = sz-1, gg = n, cz = 0;i>-1; cz = gg-g[i], gg = g[i], i--){ 
                        for(int j = 0; j < min(gg-g[i], n-gg); j++) a[ j + g[i] ] = a[ j + gg ];
                        if( gg - g[i] > n - gg && (cz == 0 || (gg - g[i]) % cz == 0) ){ 
                                int j = 0;
                                for(; j < cz; j++)
                                        if( a[j + gg-cz ] != a[ gg + j ] ) break;
                                if( j == cz ) a[gg - 1] = 1 - a[gg + cz - 1]; 
                        }
                        a[g[i]] = 0;
                }
                for(int i = 0; i < n; i++)
                        printf("%d",a[i]),a[i] = 0;
                printf("\n");
        }
}

Let's consider a₀, minimum of them.

The point is when you can make x, you also can make x + a₀.

So, try finding some of the minimum numbers you can construct in a way that you can construct all of the numbers from them.

Complexity:

//                           <................ be name khoda .................>
#include<bits/stdc++.h>
using namespace std;
int iin(){int x;scanf("%d",&x);return x;}
int n, k;
const int N = 10001;
const int maxA = 100001;
const int inf = 1e9 + 1;
int ans[ maxA + 1];
int mn = 0;
main()
{
	fill( ans, ans + maxA + 1, inf);
	n = iin();
	ans[ 0 ] = 0;
	mn = iin();
	n--;
	while(n--){
		int cc = iin();
		int gcd = __gcd(mn, cc);
		int t = mn / gcd;
		for(int j = 0;j < gcd; j++){
			int c = inf;
			for(int k = 0; k < t * 2; k++){
				int I = ( j + ((cc%mn) * k )) % mn;
				if( ans[ I ] < c ) c = ans[ I ];
				else ans[ I ] = c;
				c += cc;
			}
		}
	}
	k = iin();
	while( k-- ){
		int c = iin();
		if( ans[ c % mn ] > c ) printf("NIE\n");
		else printf("TAK\n");
	}
	
}

Actually, it uses dijkstra to implement what I said in hint 3.

Complexity:

#include <bits/stdc++.h>
using namespace std;
int INF = 2000000000,A[5005],dist[50005];
set < pair<int,int> > dijk;
int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%d", &A[i]);
	}
	for (int i = 0; i < A[0]; ++i)
	{
		dist[i] = INF;
	}
	dist[0] = 0;
	dijk.insert(make_pair(dist[0],0));
	while(!dijk.empty())
	{
		pair<int,int> top = *dijk.begin();
		dijk.erase(top);
		int v = top.second;
		int d = top.first;
		for (int i = 1; i < n; ++i)
		{
			int d2 = d+A[i];
			int v2 = (v+A[i])%A[0];
			if(d2 < dist[v2])
			{
				if(dist[v2] < INF)
					dijk.erase(make_pair(dist[v2],v2));
				dist[v2] = d2;
				dijk.insert(make_pair(dist[v2],v2));
			}
		}
	}
	int k;
	scanf("%d", &k);
	while(k--)
	{
		int x;
		scanf("%d", &x);
		if(dist[x%A[0]] <= x)
			printf("TAK\n");
		else
			printf("NIE\n");
	}
	return 0;
}

582C - Превосходящие периодические подмассивы

632E - Вор в магазине