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 mod = 1e9 + 7;
int power(int a, int b) {
  int res = 1;
  while (b) {
    if (b % 2 != 0) {
      res = (1LL * res * a) % mod;
    }
    a = (1LL * a * a) % mod;
    b /= 2;
  }
  return res;
}
int ans[5005][5005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  if (n <= 5000) {
    int answer = 0;
    for (int i = 0; i <= n; i++) {
      ans[i][0] = 1;
      for (int j = 1; j <= i; j++) {
        ans[i][j] = (ans[i - 1][j - 1] + ans[i - 1][j]) % mod;
      }
    }
    for (int i = 1; i <= n; i++) {
      answer = (answer + (1LL * ans[n][i] * power(i, k)) % mod) % mod;
    }
    printf("%d\n", answer);
  } else {
    int prev = 1;
    for (int i = 1; i <= k; i++) {
      for (int d = 0; d <= k; d++) {
        if (i == 1) {
          prev = (1LL * prev * (n - d)) % mod;
          ans[1][d] = (1LL * prev * power(2, n - d - 1)) % mod;
        } else {
          ans[i][d] =
              ((1LL * (d + 1) * ans[i - 1][d]) % mod + ans[i - 1][d + 1]) % mod;
        }
      }
    }
    printf("%d\n", ans[k][0]);
  }
  return 0;
}
