I started by counting solutions with subsegments not including ones, then includeing 1s and not including 2s then includeing 2s and 1s not including 3s ... I used combination for counting this stuff which you can make O(1) easily, so my approach is O(n) which passes(and passed).

// time-limit: 2000
// problem-url: https://codeforces.me/contest/1793/problem/D
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx2,tune=native")
#define int int64_t
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define txtio(s) freopen((string(s)+string(".in")).c_str(),"r",stdin);freopen((string(s)+string(".out")).c_str(),"w",stdout);
#define F first
#define S second
#define MP make_pair
#define PB push_back
#define sqr(a) (a)*(a)
#define endl '\n'
#define I insert
const int mod=1000000007;
int binpow(int a, int b) {
	int res=1;
	while (b>0) {
		if (b&1)
			res=res*a;
		a=a*a;
		b>>=1;
	}
	return res;
}
void solve(int wasd){
	//cout<<"Case "<<wasd<<" :";
	int n;
	cin>>n;
	vector<int> p(n,0),q(n,0);
	for(int i=0; i<n; i++){
		int tmp;
		cin>>tmp;
		p[tmp-1]=i;
	}
	for(int i=0; i<n; i++){
		int tmp;
		cin>>tmp;
		q[tmp-1]=i;
	}
	int ans=(abs(p[0]-q[0])-1)*(abs(p[0]-q[0])-2)/2+min(p[0],q[0])*(min(p[0],q[0])-1)/2+(n-max(p[0],q[0])-1)*(n-2-max(p[0],q[0]))/2+n-max(p[0],q[0])-1+min(p[0],q[0])+abs(p[0]-q[0])-1;
	int curl=min(p[0],q[0]),curr=max(p[0],q[0]);
	for(int i=1; i<n; i++){
		if(curl==0&&curr==n-1){
			break;
		}
		if((p[i]>=curl&&p[i]<=curr)||(q[i]>=curl&&q[i]<=curr)){
			curl=min(curl,min(p[i],q[i]));curr=max(curr,max(p[i],q[i]));
			continue;
		}
		if(min(p[i],q[i])>curr){
			ans+=(curl+1)*(min(p[i],q[i])-curr);
		}
		else if(max(p[i],q[i])<curl){
			ans+=(curl-max(p[i],q[i]))*(n-curr);
		}
		else ans+=(max(p[i],q[i])-curr)*(curl-min(p[i],q[i]));
		curl=min(curl,min(p[i],q[i]));curr=max(curr,max(p[i],q[i]));
	}
	ans++;//solution where all of the array is included
	cout<<ans<<endl;
}
int32_t main(){
	fastio;
	int t=1;
	//cin>>t;
	for(int i=0; i<t; i++) solve(i);
}