#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = (a), i##ABRACADABRA = (b); i <= i##ABRACADABRA; i++)
#define drep(i, a, b) for (int i = (a), i##ABRACADABRA = (b); i >= i##ABRACADABRA; i--)
using namespace std;
using ll = long long;

int n,k,a[222];

int main() {
  scanf("%d%d",&n,&k);
  rep(i,1,n)a[i]=n-1;
  puts("Second"),fflush(stdout);
  mt19937 rng(random_device{}());
  while (1){
    int o,z;
    scanf("%d%d",&o,&z);
    if (o<=0)break;
    a[o]-=z;
    vector<array<int,2>>vec;
    rep(i,1,n)if (a[i]>0)vec.push_back({a[i],i});
    sort(vec.rbegin(),vec.rend());
    if ((int)vec.size()>z)vec.resize(z);
    printf("%d\n",(int)vec.size()),fflush(stdout);
    for (auto [x,y]:vec)printf("%d ",y),--a[y];
    puts(""),fflush(stdout);
  }
  return 0;
}

After trying $$$K=1$$$, we take a leap of faith and extrapolate to guess that Second wins every time.

Let us experiment with which positions are "better" for Second and which are "better" for First. One direct winning position for First that we can construct is one tall pile of stones. If we "flip" this configuration, having many piles of stones of size $$$1$$$, Second wins.

This suggests that it is beneficial to Second to have a large number of piles of stones, with the piles themselves being very short. Thus, we will try to remove stones from the $$$K$$$ piles with the largest number.

Note that unlike the above solution by L_Wave, this fails since we remove $$$M$$$ stones rather than $$$j$$$ stones (where $$$j$$$ is the number taken by First). This solution ends up removing piles of $$$1$$$ stone each, and such piles are beneficial for us (e.g. $$$K=2$$$, pile sizes are $$$[2,1]$$$, then this strategy makes the pile sizes $$$[1,0]$$$ and allows First to win). Failed submission

To address this, we only remove stones from piles with $$$M$$$ stones, where $$$M$$$ is the maximum number of stones in any pile. Intuitively, this allows the size of the piles to decrease faster than First decreases the number of piles.

However, this reasoning is somewhat bogus because the solution is incorrect. Take the following example with $$$N=4$$$ and $$$K=2$$$.

Where $\rightarrow^1$ and $$$\rightarrow^2$$$ represent moves by First and Second respectively.

Submission

We play as the 2nd player, removing stones from the largest $$$K$$$ piles (all non-empty piles if there are $$$\lt K$$$ of them) with two exceptions. This aims to preserve the invariant "the number of piles is greater than the size of largest pile" at the beginning of 1st player's moves.

Special case 1: If the largest pile has more than one stone, we don't remove stones from piles with size 1 — it doesn't benefit us. In other words we don't reduce the number of piles if we don't have to, and when we do, it's a trivial winning state already.

Special case 2: We avoid moving to a state "$$$K+1$$$ piles with size $$$\gt K$$$ and one pile with size $$$\le K$$$" from a state "$$$K+2$$$ piles, all with size $$$\gt K$$$". For $$$K=1$$$ we don't have to worry about it. Since we were trying to remove $$$K$$$ stones, such a move can be fixed by removing one less stone, and as long as the invariant held before, it also means the fixed move gives $$$K+2$$$ piles with size $$$K+1$$$ each.