#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <utility>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <numeric>
#include <functional>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define exrep(i, a, b) for(ll i = a; i <= b; i++)
#define out(x) cout << x << endl
#define exout(x) printf("%.10f\n", x)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define pb push_back
#define re0 return 0
const ll mod = 1000000007;
const ll INF = 1e16;

const ll MAX = 100010;

// a^n (mod.MOD)を求める。計算量はO(logn)
ll modpow(ll a, ll n, ll MOD = mod) {
    if(n == 0) {
        return 1;
    }
    if(n%2 == 1) {
        return (a * modpow(a, n-1, MOD)) % MOD;
    }
    return (modpow(a, n/2, MOD) * modpow(a, n/2, MOD)) % MOD;
}

ll inverse(ll a) {
    return modpow(a, mod - 2);
}

ll fact[MAX];  // fact[i] : iの階乗のmod
ll invfact[MAX];  // invfact[i] : iの階乗の逆数のmod

void init() {
    fact[0] = invfact[0] = 1;
    for(ll i = 1; i < MAX; i++) {
        fact[i] = (i * fact[i-1]) % mod;
        invfact[i] = inverse(fact[i]);
    }
}

// nCrをO(n)で求める。 
ll Comb(ll n, ll r) {
    if(r < 0 || n < 0 || n < r) {
        return 0;
    }
    ll res = fact[n];
    res = (res * invfact[r]) % mod;
    res = (res * invfact[n-r]) % mod;
    return res;
}

int main() {
    ll n, m;
    cin >> n >> m;

    init();

    ll ans = 0;
    exrep(i, 1, m) {
        ll x = Comb(m, i) * modpow(i, n) % mod;
        if(i % 2 == m % 2) {
            ans = (ans + x) % mod;
        }
        else {
            ans = (ans + mod - x) % mod;
        }
    }

    out(ans);
    re0;
}