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;
int const p = 1e9 + 7;
int pw(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % p;
    x = 1ll * x * x % p;
    y >>= 1;
  }
  return res;
}
int s[5005][5005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  s[1][1] = 1;
  for (int i = 2; i <= k; i++)
    for (int j = 1; j <= i; j++)
      s[i][j] = (s[i - 1][j - 1] + 1ll * j * s[i - 1][j]) % p;
  int t = 1, ans = 0;
  for (int i = 0; i <= min(n, k); i++)
    ans += 1ll * t * s[k][i] % p * pw(2, n - i) % p, ans %= p,
        t = 1ll * t * (n - i) % p;
  printf("%d", ans);
  return 0;
}
