Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mo = 1e9 + 7;
const int inv2 = (mo + 1) >> 1;
const int N = 5005;
int n, k, S2[N][N];
int qpow(int a, int b) {
  int ans = 1;
  for (; b >= 1; b >>= 1, a = 1ll * a * a % mo)
    if (b & 1) ans = 1ll * ans * a % mo;
  return ans;
}
int main() {
  S2[0][0] = 1;
  n = N - 5;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      S2[i][j] = (S2[i - 1][j - 1] + 1ll * j * S2[i - 1][j]) % mo;
    }
  }
  scanf("%d%d", &n, &k);
  int ans = 0, bin = qpow(2, n), tmp = 1;
  for (int i = 1; i <= k && i <= n; i++) {
    bin = 1ll * bin * inv2 % mo;
    tmp = 1ll * tmp * (n - i + 1) % mo;
    int r = 1ll * S2[k][i] * tmp % mo * bin % mo;
    (ans += r) >= mo ? ans -= mo : 0;
  }
  printf("%d\n", ans);
  return 0;
}