#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
//#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = ll(A); I < ll(B); ++I)
#define FORR(I,A,B) for(ll I = ll((B)-1); I >= ll(A); --I)
#define TO(x,t,f) ((x)?(t):(f))
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin()) //xi>=v  x is sorted
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin()) //xi>v  x is sorted
#define NUM(x,v) (POSU(x,v)-POSL(x,v))  //x is sorted
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd(ll a,ll b){if(a%b==0)return b;return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}
#define NEXTP(x) next_permutation(x.begin(),x.end())
const ll INF=ll(1e16)+ll(7);
const ll MOD=1000000007LL;
#define out(a) cout<<fixed<<setprecision((a))
//tie(a,b,c) = make_tuple(10,9,87);
#define pop(a) __builtin_popcount((a))
ll keta(ll a){ll r=0;while(a){a/=10;r++;}return r;}



ll cnt[52][52];
ll B[52][52];
ll M;
vector< vector<ll> > pow_(vector<vector<ll>> A,ll k){
  vector< vector<ll> > res(M,vector<ll>(M,0));
  if(k==0){
    FOR(i,0,M)res[i][i] = 1;
    return res;
  }
  if(k==1){
    return A;
  }
  auto x = pow_(A,k/2);

  FOR(i,0,M){
    FOR(j,0,M){
      FOR(m,0,M){
        (res[i][j] += x[i][m]*x[m][j]) %=MOD;
      }
    }
  }

  if(k&1){
    FOR(i,0,M)FOR(j,0,M)x[i][j] = 0;
    FOR(i,0,M){
      FOR(j,0,M){
        FOR(m,0,M){
          (x[i][j] += res[i][m]*B[m][j]) %=MOD;
        }
      }
    }
    return x;
  }
  return res;
}




int main(){


  ll K;
  cin >> M >> K;

  FOR(i,0,M){
    FOR(j,0,M){
      cnt[i][(i*j)%M]++;
      cnt[i][(i+j)%M]++;
    }
  }

  vector< vector<ll> > A(M,vector<ll>(M,0));
  FOR(i,0,M)FOR(j,0,M){
    A[i][j] = cnt[j][i];
    B[i][j] = cnt[j][i];
  }

  auto ans = pow_(A,K);
  cout << ans[0][0] << endl;
}