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 i, j, k, n, x, y, f, s, t;
long long a[5010][5010], ans, temp, q = 1e9 + 7;
long long work(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2 == 1) z = z * x % q;
    x = x * x % q;
    y = y / 2;
  }
  return z;
}
int main() {
  scanf("%d%d", &n, &k);
  ans = 0;
  a[0][0] = 1;
  for (i = 0; i < k; i++) {
    for (j = 0; j <= k; j++) {
      a[i + 1][j] += j * a[i][j] % q;
      a[i + 1][j + 1] += (n - j) * a[i][j] % q;
    }
  }
  for (i = 0; i <= k; i++) {
    temp = a[k][i] * work(2, n - i) % q;
    ans = (ans + temp) % q;
  }
  cout << ans << endl;
}
