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 MAXN = 5010;
int n, K, S[MAXN][MAXN];
const int mod = int(1e9) + 7;
long long power(long long e, long long x, int m = mod) {
  long long ret = 1;
  for (; x; x >>= 1, (e *= e) %= mod)
    if (x & 1) (ret *= e) %= mod;
  return ret;
}
int main() {
  scanf("%d%d", &n, &K);
  S[0][0] = 1;
  for (int i = 1; i <= K; ++i)
    for (int j = 0; j <= i; ++j) {
      (S[i][j] += (long long)S[i - 1][j - 1] * (n - j + 1) % mod) %= mod;
      (S[i][j] += (long long)S[i - 1][j] * j % mod) %= mod;
    }
  int ans = 0;
  for (int i = 1; i <= K && i <= n; ++i)
    (ans += power(2, n - i) * S[K][i] % mod) %= mod;
  printf("%d\n", ans);
}
