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 std::max;
using std::min;
using std::swap;
using std::vector;
const int N = 5005;
const int mod = 1e9 + 7;
int n, k, s[N][N];
int ksm(int a, int b, int ans = 1) {
  while (b) {
    if (b & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return ans;
}
int getint() {
  int X = 0, w = 0;
  char ch = getchar();
  while (!isdigit(ch)) w |= ch == '-', ch = getchar();
  while (isdigit(ch)) X = X * 10 + ch - 48, ch = getchar();
  if (w) return -X;
  return X;
}
signed main() {
  n = getint(), k = getint();
  s[0][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++)
      s[i][j] = (s[i - 1][j - 1] + 1ll * s[i - 1][j] * j % mod) % mod;
  int ans = 0, now = n, in = ksm(2, mod - 2), now0 = ksm(2, n - 1);
  for (int j = 1; j <= k; j++) {
    ans = (ans + 1ll * s[k][j] * now % mod * now0 % mod) % mod;
    now = 1ll * now * (n - j) % mod;
    now0 = 1ll * now0 * in % mod;
  }
  printf("%d\n", ans);
  return 0;
}
