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 = 5005;
const long long mod = 1e9 + 7;
int n, k;
int s[maxn][maxn];
int qpow(int a, int b) {
  if (b < 0) return 0;
  int ans = 1;
  a %= mod;
  while (b) {
    if (b & 1) ans = (1ll * ans * a) % mod;
    b >>= 1, a = (1ll * a * a) % mod;
  }
  return ans % mod;
}
void init() {
  memset(s, 0, sizeof(s));
  s[0][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++)
      s[i][j] = (1ll * j * s[i - 1][j] % mod + s[i - 1][j - 1]) % mod;
}
int main() {
  int ans = 0;
  int now = 1;
  scanf("%d%d", &n, &k);
  init();
  for (int j = 0; j <= k; j++) {
    ans = (ans + 1ll * s[k][j] * qpow(2, n - j) % mod * now % mod) % mod;
    now = (1ll * now * (n - j)) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
