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;
int n, k;
long long ans, s2[5001][5001], jc = 1;
long long ksm(int b) {
  long long an = 1, a = 2;
  while (b) {
    if (b & 1) an = (an * a) % (1000000007);
    a = (a * a) % (1000000007);
    b >>= 1;
  }
  return an % (1000000007);
}
int main() {
  scanf("%d%d", &n, &k);
  s2[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      s2[i][j] =
          (s2[i - 1][j - 1] + (j * s2[i - 1][j]) % (1000000007)) % (1000000007);
    }
  }
  long long kk = ksm(n), nw = ksm((1000000007) - 2);
  for (int i = 0; i <= k; i++) {
    ans = (ans + ((((s2[k][i] * kk)) % (1000000007)) * jc) % (1000000007)) %
          (1000000007);
    kk = (kk * nw) % (1000000007);
    jc = (jc * (n - i)) % (1000000007);
  }
  printf("%lld\n", ans);
  return 0;
}
