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 = 5000 + 10, md = 1e9 + 7;
int n, k;
int s[N][N];
int qpow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % md)
    if (b & 1) res = 1ll * res * a % md;
  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 * s[i - 1][j] * j % md) % md;
    }
  }
  int res = 0;
  int fac = 1;
  for (int i = 1; i <= min(n, k); i++) {
    fac = 1ll * fac * ((n - i + 1) % md) % md;
    int tmp = 1ll * s[k][i] * fac % md * qpow(2, n - i) % md;
    res = (res + tmp) % md;
  }
  printf("%d\n", res);
  return 0;
}