#include<bits/stdc++.h>

using namespace std;

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=n-1;i>=a;i--)

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define SZ(x) ((int)(x).size())

typedef vector<int> VI;

typedef long long ll;

typedef pair<int,int> PII;

const ll mod=;

ll powmod(ll a,ll b) {ll res=;a%=mod; assert(b>=); for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}

// head

namespace linear_seq {

    const int N=;

    ll res[N],base[N],_c[N],_md[N];

    vector<int> Md;

    void mul(ll *a,ll *b,int k) {

        rep(i,,k+k) _c[i]=;

        rep(i,,k) if (a[i]) rep(j,,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;

        for (int i=k+k-;i>=k;i--) if (_c[i])

            rep(j,,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;

        rep(i,,k) a[i]=_c[i];

    }

    int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...

//        printf("SIZE %d\n",SZ(b));

        ll ans=,pnt=;

        int k=SZ(a);

        assert(SZ(a)==SZ(b));

        rep(i,,k) _md[k--i]=-a[i];_md[k]=;

        Md.clear();

        rep(i,,k) if (_md[i]!=) Md.push_back(i);

        rep(i,,k) res[i]=base[i]=;

        res[]=;

        while ((1ll<<pnt)<=n) pnt++;

        for (int p=pnt;p>=;p--) {

            mul(res,res,k);

            if ((n>>p)&) {

                for (int i=k-;i>=;i--) res[i+]=res[i];res[]=;

                rep(j,,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;

            }

        }

        rep(i,,k) ans=(ans+res[i]*b[i])%mod;

        if (ans<) ans+=mod;

        return ans;

    }

    VI BM(VI s) {

        VI C(,),B(,);

        int L=,m=,b=;

        rep(n,,SZ(s)) {

            ll d=;

            rep(i,,L+) d=(d+(ll)C[i]*s[n-i])%mod;

            if (d==) ++m;

            else if (*L<=n) {

                VI T=C;

                ll c=mod-d*powmod(b,mod-)%mod;

                while (SZ(C)<SZ(B)+m) C.pb();

                rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;

                L=n+-L; B=T; b=d; m=;

            } else {

                ll c=mod-d*powmod(b,mod-)%mod;

                while (SZ(C)<SZ(B)+m) C.pb();

                rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;

                ++m;

            }

        }

        return C;

    }

    int gao(VI a,ll n) {

        VI c=BM(a);

        c.erase(c.begin());

        rep(i,,SZ(c)) c[i]=(mod-c[i])%mod;

        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));

    }

};

int a[][];

long long det(int n) {

    long long ans = ;

    for (int i = ; i < n; i++) {

        for (int j = i + ; j < n; j++) {

            while (a[j][i] != ) {

                int u = a[i][i] / a[j][i];

                for (int k = ; k < n; k++) {

                    int t = (a[i][k] - (long long)a[j][k] * u % mod + mod) % mod;

                    a[i][k] = a[j][k];

                    a[j][k] = t;

                }

                ans = -ans;

            }

        }

        ans = ans * a[i][i] % mod;

    }

    if (ans < ) {

        ans += mod;

    }

    return ans;

}

long long work(int k, int n) {

//  assert(n > 2 * k + 1);

    memset(a, , sizeof a);

    for (int i = ; i < n; i++) {

        a[i][i] = k;

        for (int j = ; j <= k; j++) {

            a[i][(i + j) % n] = -;

        }

    }

/*

    for (int i = 0; i < n; i++) {

        for (int j = 0; j < n; j++) {

            printf("%d%c", a[i][j], j == n - 1 ? '\n' : ' ');

        }

    }

*/

    int t = ;

    for (int i = ; i < k; i++) {

        t = t * i % mod;

    }

    return (long long)det(n - ) * powmod(t, n) % mod;

}

void maketable() {

    printf("{\n");

    for (int k = ; k <= ; k++) {

        if (k > ) {

            printf(",\n");

        }

        printf("{\n");

        for (int i =  * k + ; i <= ; i++) {

            if (i >  * k + ) {

                printf(",");

            }

            printf("%lld", work(k, i));

        }

        printf("}\n");

    }

    printf("}\n");

}

int main() {

    int k;

    long long n;

//  maketable();

    cin >> k >> n;

//  k = 5;

//  printf("%d\n", work(k, n));

    vector<int> a;

    for (int i =  * k + ; i <= ( << k) +  * k + ; i++) {

//      printf("%d %d\n", i, work(k, i));

        a.push_back(work(k, i));

    }

//  cout << linear_seq::gao(b[k], n - (2 * k + 1)) << endl;

    cout << linear_seq::gao(a, n - ( * k + )) << endl;

    return ;

}