#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long ll; typedef vector vl; typedef vector> vvl; typedef pair 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_K = 200010; // 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_K]; // fact[i] : iの階乗のmod ll invfact[MAX_K]; void init() { fact[0] = invfact[0] = 1; for(ll i = 1; i < MAX_K; i++) { fact[i] = (i*fact[i-1])%mod; invfact[i] = inverse(fact[i]); } } 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; } // 区別できるn個のものを区別できるkグループに分類する方法の数をO(k*log(n))で求める int main() { ll n, k; cin >> n >> k; init(); ll X = 0; ll Y = 0; ll parity = 0; for(ll i = k; i >= 1; i--) { ll A = Comb(k, i); ll B = modpow(i, n); if(parity%2 == 0) { X += (A*B)%mod; X %= mod; } else { Y += (A*B)%mod; Y %= mod; } parity++; } ll ans = (X + mod - Y)%mod; out(ans); re0; }