Problem: You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk. 

Output the sum of costs over all non-empty subsets of people.

Input

Only line of input contains two integers N (1 ≤ N ≤ 109) representing total number of people and k (1 ≤ k ≤ 5000).

Output

Output the sum of costs for all non empty subsets modulo 109 + 7.

Examples

Input

1 1


Output

1


Input

3 2


Output

24

Note

In the first example, there is only one non-empty subset {1} with cost 11 = 1.

In the second example, there are seven non-empty subsets.

- {1} with cost 12 = 1

- {2} with cost 12 = 1

- {1, 2} with cost 22 = 4

- {3} with cost 12 = 1

- {1, 3} with cost 22 = 4

- {2, 3} with cost 22 = 4

- {1, 2, 3} with cost 32 = 9

The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 5010;
const int mod = 1e9 + 7;
int jc[N], n, K, S[N][N], C[N], pw[N], ny[N];
int ksm(int x, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    k >>= 1;
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &K);
  S[0][0] = 1;
  for (int i = 1; i <= K; i++) {
    for (int j = 1; j <= i; j++)
      (S[i][j] = (S[i - 1][j - 1] + 1ll * j * S[i - 1][j] % mod) % mod) %= mod;
  }
  ny[1] = ny[0] = jc[0] = 1;
  for (int i = 1; i <= K; i++) jc[i] = 1ll * jc[i - 1] * i % mod;
  for (int i = 2; i <= K; i++)
    ny[i] = (mod - 1ll * (mod / i) * ny[mod % i] % mod) % mod;
  pw[0] = ksm(2, n);
  for (int i = 1; i <= K; i++) pw[i] = 500000004ll * pw[i - 1] % mod;
  C[0] = 1;
  for (int i = 1; i <= K; i++) {
    if (!C[i - 1]) continue;
    C[i] = 1ll * C[i - 1] * (n - i + 1) % mod * ny[i] % mod;
  }
  int res = 0;
  for (int i = 0; i <= K; i++)
    (res += 1ll * S[K][i] * jc[i] % mod * C[i] % mod * pw[i] % mod) %= mod;
  printf("%d", res);
  return 0;
}