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.
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int n, k, S[5005][5005], ans = 0;
int Power(int x, int y, int mod) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % mod;
    x = 1ll * x * x % mod, y >>= 1;
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &k);
  S[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= k; j++) {
      S[i][j] = (S[i - 1][j - 1] + 1ll * S[i - 1][j] * j) % mod;
    }
  }
  for (int i = 0; i <= k && i <= n; i++) {
    int tmp = 1;
    for (int j = n; j > n - i; j--) tmp = 1ll * tmp * j % mod;
    ans = (ans + 1ll * tmp * S[k][i] % mod * Power(2, n - i, mod) % mod) % mod;
  }
  printf("%d\n", ans);
}
