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 Maxk = 5005;
const int md = 1e9 + 7;
int n, k, m;
long long dp[Maxk][Maxk];
long long fpow(int num) {
  long long rez = 1;
  if (num == 0) return rez;
  rez = fpow(num / 2);
  rez = (rez * rez) % md;
  if (num & 1) rez = (rez * 2) % md;
  return rez;
}
int main() {
  scanf("%d%d", &n, &k);
  m = min(n, k);
  dp[0][m] = fpow(n - m);
  for (int j = m - 1; j >= 0; j--) {
    dp[0][j] = (dp[0][j + 1] * 2) % md;
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j <= m; j++) {
      dp[i][j] = (j * dp[i - 1][j] + (n - j) * dp[i - 1][j + 1]) % md;
    }
  }
  printf("%d\n", dp[k][0]);
  return 0;
}
