#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
void RADD(int &a, int b) {
if ((a += b) >= MOD) a -= MOD;
if (a < 0) a += MOD;
}
int MUL(int a, int b) {
return 1ll * a * b % MOD;
}
int PushDP(int N, int K) { // O(N^2) DP solution
vector<vector<int>> dp(N + 1, vector<int>(K + 2, 0));
dp[0][0] = 1; // dp[i][j] = answer for N = i and K = j
for (int i = 0; i < N; i++) {
for (int j = 0; j <= K; j++) {
// There are j positions where we insert at back, K doesn't increase
RADD(dp[i + 1][j], MUL(dp[i][j], j));
// Among i+1 possible places to be inserted, inserting other than the back increases K,
// of which there are (i + 1 - j) positions.
RADD(dp[i + 1][j + 1], MUL(dp[i][j], i + 1 - j));
}
}
return dp[N][K];
}
int PullDP(int N, int K) { // O(N^2) DP solution
vector<vector<int>> dp(N + 1, vector<int>(K + 2, 0));
dp[0][0] = 1; // dp[i][j] = answer for N = i and K = j
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= K; j++) {
RADD(dp[i][j], MUL(dp[i - 1][j], j));
RADD(dp[i][j], MUL(dp[i - 1][j - 1], i + 1 - j));
}
}
return dp[N][K];
}
int Power(int n, int x) {
if (x == 0) return 1;
int res = Power(n, x / 2);
res = MUL(res, res);
if (x & 1) res = MUL(res, n);
return res;
}
int Solve(int N, int K) { // O(N log N) combinatorial solution
vector<int> fact(N + 2);
vector<int> invfact(N + 2);
invfact[0] = fact[0] = 1;
for (int i = 1; i <= N + 1; i++) {
fact[i] = MUL(i, fact[i - 1]);
invfact[i] = Power(fact[i], MOD - 2); // by Fermat's Little Theorem
}
auto C = [&](int n, int k) { // binomial
if (k < 0 || k > n) return 0;
return MUL(fact[n], MUL(invfact[k], invfact[n - k]));
};
int ans = 0;
// sum_{k = 0}^{K}((-1)^k * C(N + 1, k) * (K - k + 1)^N)
for (int k = 0, sgn = 1; k < K; k++, sgn *= -1) {
RADD(ans, sgn * MUL(C(N + 1, k), Power(K - k, N)));
}
return ans;
}
int main() {
int N, K;
cin >> N >> K;
cout << Solve(N, K) << "\n";
return 0;
}
