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 n, k;
int S[5005][5005], fac[5005], ans;
int ksm(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % 1000000007;
    x = 1ll * x * x % 1000000007, y /= 2;
  }
  return res;
}
int C(int x, int y) {
  int res = ksm(fac[y], 1000000007 - 2);
  for (int i = x - y + 1; i <= x; i++) res = 1ll * res * i % 1000000007;
  return res;
}
int main() {
  scanf("%d %d", &n, &k);
  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 * j * S[i - 1][j] % 1000000007) % 1000000007;
  }
  fac[0] = 1;
  for (int i = 1; i <= k; i++) fac[i] = 1ll * i * fac[i - 1] % 1000000007;
  for (int i = 1; i <= k && i <= n; i++)
    (ans += 1ll * C(n, i) * S[k][i] % 1000000007 * fac[i] % 1000000007 *
            ksm(2, n - i) % 1000000007) %= 1000000007;
  printf("%d\n", ans);
  return 0;
}
