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 P = 1e9 + 7;
int C[5000 + 10][5000 + 10];
int fpow(int a, int b) {
  int c = 1;
  while (b) {
    if (b & 1) c = (long long)c * a % P;
    b >>= 1;
    a = (long long)a * a % P;
  }
  return c;
}
int a[5000 + 10][5000 + 10];
int cal(int n, int k) {
  int ret = fpow(2, n - k);
  for (int i = 0; i < k; i++) ret = (long long)ret * (n - i) % P;
  return ret;
}
int main() {
  int n, k;
  cin >> n >> k;
  int ans = 0;
  if (n < k) {
    for (int i = 0; i <= k; i++) {
      C[i][0] = 1;
      for (int j = 1; j <= i; j++)
        C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
    }
    for (int i = 1; i <= n; i++)
      ans = (ans + (long long)fpow(i, k) * C[n][i]) % P;
  } else {
    a[0][0] = 1;
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++)
        a[i][j] = ((long long)j * a[i - 1][j] + a[i - 1][j - 1]) % P;
    for (int i = 1; i <= k; i++)
      ans = (ans + (long long)a[k][i] * cal(n, i)) % P;
  }
  printf("%d\n", ans);
  return 0;
}
