#include <bits/stdc++.h>
// clang-format off
#include <atcoder/all>
using namespace std;
using namespace atcoder;
#define rep(i,n) for(int i=0;i<(n);i++)
template<class T,class S>inline bool chmax(T &a,const S &b) {return (a<b? a=b,1:0);}
template<class T,class S>inline bool chmin(T &a,const S &b) {return (a>b? a=b,1:0);}
template<class T,class S>inline T floor(T a,S b) {if(b<0)a=-a,b=-b;return a>=0?a/b:(a+1)/b-1;}
template<class T,class S>inline T ceil(T a,S b) {if(b<0)a=-a,b=-b;return a>0?(a-1)/b+1:a/b;}
constexpr int dx[8]={1,0,-1,0,1,-1,1,-1},dy[8]={0,1,0,-1,1,-1,-1,1};
using ll = long long;
// clang-format on
#ifdef LOCAL
#include <util/debug_print.hpp>
#else
#define debug(...) static_cast<void>(0)
#endif

using mint = modint1000000007;
ostream &operator<<(ostream &os, const mint &a) {
    return os << a.val();
}

/*
 modint 上のCombination計算を行う構造体

 計算量
    初期化：O(√n)
    計算 : O(1)
*/
struct combination {
    vector<mint> fact, ifact;
    static const int _MAX = 5100000;
    combination(int n = _MAX) : fact(n + 1), ifact(n + 1) {
        assert(n <= _MAX);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
    }
    mint operator()(int n, int k) {
        if (n == 0 && k == 0) return 1;
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n, m;
    cin >> n >> m;
    combination cmb;

    // 制約をやぶる箱の数
    mint ans = 0;
    for (int i = 0; i <= m; i++) {
        mint x = cmb(m, i) * mint(m - i).pow(n);
        ans += (i % 2 == 0 ? 1 : -1) * x;
    }
    cout << ans << '\n';
    return 0;
}