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;
long long mo = (long long)1e9 + 7;
long long a[5005][5005], n, k;
long long fpow(long long a, long long b, long long c) {
  if (b < 0) return 0;
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    b >>= 1;
    a = a * a % c;
  }
  return ans;
}
int main() {
  while (~scanf("%lld%lld", &n, &k)) {
    memset(a, 0, sizeof(a));
    a[0][0] = 1;
    for (int i = 1; i <= k; i++) {
      a[i][0] = 0;
      for (int j = 1; j <= i - 1; j++) {
        a[i][j] = j * a[i - 1][j] % mo;
        if (n - j + 1 > 0)
          a[i][j] = (a[i][j] + (n - j + 1) * a[i - 1][j - 1] % mo) % mo;
      }
      if (n - i + 1 > 0) a[i][i] = a[i - 1][i - 1] * (n - i + 1) % mo;
    }
    long long ans = 0;
    for (int i = 1; i <= k; i++) {
      ans = (ans + fpow(2, n - i, mo) * a[k][i] % mo) % mo;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
