#include "bits/stdc++.h" using namespace std; #ifdef _DEBUG #include "dump.hpp" #else #define dump(...) #endif //#define int long long #define rep(i,a,b) for(int i=(a);i<(b);i++) #define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--) #define all(c) begin(c),end(c) const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f; const int MOD = 1'000'000'007; template bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; } template bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; } template struct ModInt { static const int kMod = MOD; unsigned x; ModInt() :x(0) {} ModInt(signed x_) { x_ %= MOD; if (x_ < 0)x_ += MOD; x = x_; } ModInt(signed long long x_) { x_ %= MOD; if (x_ < 0)x_ += MOD; x = x_; } int get()const { return (int)x; } ModInt &operator+=(ModInt m) { if ((x += m.x) >= MOD)x -= MOD; return *this; } ModInt &operator-=(ModInt m) { if ((x += MOD - m.x) >= MOD)x -= MOD; return *this; } ModInt &operator*=(ModInt m) { x = (unsigned long long)x*m.x%MOD; return *this; } ModInt &operator/=(ModInt m) { return *this *= m.inverse(); } ModInt operator+(ModInt m)const { return ModInt(*this) += m; } ModInt operator-(ModInt m)const { return ModInt(*this) -= m; } ModInt operator*(ModInt m)const { return ModInt(*this) *= m; } ModInt operator/(ModInt m)const { return ModInt(*this) /= m; } ModInt operator-()const { return ModInt(MOD - x); } bool operator==(ModInt m)const { return x == m.x; } bool operator!=(ModInt m)const { return x != m.x; } ModInt inverse()const { signed a = x, b = MOD, u = 1, v = 0; while (b) { signed t = a / b; a -= t * b; swap(a, b); u -= t * v; swap(u, v); } if (u < 0)u += MOD; return ModInt(u); } }; template ostream &operator<<(ostream &os, const ModInt &m) { return os << m.x; } template istream &operator>>(istream &is, ModInt &m) { signed long long s; is >> s; m = ModInt(s); return is; }; template ModInt pow(ModInt a, unsigned long long k) { ModInt r = 1; while (k) { if (k & 1)r *= a; a *= a; k >>= 1; } return r; } using mint = ModInt; // スターリング数 // O(n^2) // Verified: // https://yukicoder.me/submissions/330289 vector> stirlingNumbers(int n) { vector> S(n + 1, vector(n + 1, 0)); S[0][0] = 1; for (int i = 0; i < n; i++) { for (int j = 0; j <= i; j++) { S[i + 1][j] += S[i][j] * j; S[i + 1][j + 1] += S[i][j]; } } return S; } // 完全2部グラフの彩色多項式 // O(n^2 + n lg m) mint chromaticPolynomialOfCompleteBipartiteGraph(int n, int m, mint x) { auto S = stirlingNumbers(n); mint ret = 0; mint a = x; for (int k = 1; k <= n; k++) { ret += S[n][k] * a * pow(x - k, m); a *= x - k; } return ret; } signed main() { cin.tie(0); ios::sync_with_stdio(false); int N, M; cin >> N >> M; cout << chromaticPolynomialOfCompleteBipartiteGraph(N, M, mint(5) / 2) << endl; return 0; }