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 f[5010][5010];
int n, k;
const int mod = 1e9 + 7;
int fast_pow(int x, int a) {
  int ans = 1;
  for (; a; a >>= 1, x = 1ll * x * x % mod)
    if (a & 1) ans = 1ll * ans * x % mod;
  return ans;
}
int dp(int a, int b) {
  if (f[a][b] >= 0) return f[a][b];
  int c = n - b;
  if (a == 0) {
    f[a][b] = fast_pow(2, c);
    return f[a][b];
  }
  f[a][b] = ((b ? 1ll * b * dp(a - 1, b) : 0) +
             (c ? 1ll * c * dp(a - 1, b + 1) : 0)) %
            mod;
  return f[a][b];
}
int main() {
  scanf("%d%d", &n, &k);
  memset(f, -1, sizeof(f));
  printf("%d", dp(k, 0));
}
