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>
const int mod = 1e9 + 7;
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int qp(int d, int k) {
  int f = 1;
  while (k) {
    if (k & 1) f = (1ll * (f) * (d) % mod);
    d = (1ll * (d) * (d) % mod), k >>= 1;
  }
  return f;
}
int str[5010][5010], n, k, ans;
int main() {
  scanf("%d%d", &n, &k);
  str[0][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++)
      str[i][j] = add(str[i - 1][j - 1], (1ll * (str[i - 1][j]) * (j) % mod));
  for (int i = 1, f = 1; i <= n && i <= k; i++) {
    f = (1ll * (f) * (n - i + 1) % mod);
    ans = add(ans,
              (1ll * (f) * ((1ll * (str[k][i]) * (qp(2, n - i)) % mod)) % mod));
  }
  printf("%d\n", ans);
  return 0;
}
