Problem: There are well-known formulas: <image>, <image>, <image>. Also mathematicians found similar formulas for higher degrees.

Find the value of the sum <image> modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).

Input

The only line contains two integers n, k (1 ≤ n ≤ 109, 0 ≤ k ≤ 106).

Output

Print the only integer a — the remainder after dividing the value of the sum by the value 109 + 7.

Examples

Input

4 1


Output

10


Input

4 2


Output

30


Input

4 3


Output

100


Input

4 0


Output

4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int K = 1e6 + 6;
const int mod = 1e9 + 7;
int n, k;
int val[K];
int ans, cur;
int power(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) {
      res = (1LL * res * a) % mod;
    }
    a = (1LL * a * a) % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= k + 2; ++i) {
    val[i] = (val[i - 1] + power(i, k)) % mod;
  }
  if (n <= k + 2) {
    printf("%d\n", val[n]);
    return 0;
  }
  ans = 0;
  cur = 1;
  for (int i = 2; i <= k + 2; ++i) {
    cur = (1LL * cur * (n - i)) % mod;
    cur = (1LL * cur * power(mod + 1 - i, mod - 2)) % mod;
  }
  for (int i = 1; i <= k + 2; ++i) {
    ans = (ans + 1LL * cur * val[i]) % mod;
    cur = (1LL * cur * (n - i)) % mod;
    cur = (1LL * cur * (i - k - 2 + mod)) % mod;
    cur = (1LL * cur * power((1LL * (n - i - 1) * i) % mod, mod - 2)) % mod;
  }
  printf("%d\n", ans);
}