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 P = 1e9 + 7, M = 5005;
int S[M][M];
int fast(int a, int b) {
  int sum = 1;
  while (b > 0) {
    if (b & 1) sum = 1ll * sum * a % P;
    a = 1ll * a * a % P;
    b >>= 1;
  }
  return sum;
}
void add(int &a, int b) {
  a += b;
  if (a >= P) a -= P;
}
int main() {
  int n, K;
  scanf("%d %d", &n, &K);
  S[1][1] = 1;
  for (int i = 2; i <= K; i++) {
    S[i][1] = 1;
    for (int j = 2; j <= i; j++) {
      S[i][j] = S[i - 1][j - 1] + 1ll * S[i - 1][j] * j % P;
      if (S[i][j] >= P) S[i][j] -= P;
    }
  }
  int ans = 0;
  int A = 1;
  for (int i = 1; i <= K; i++) {
    A = 1ll * A * (n - i + 1) % P;
    add(ans, 1ll * S[K][i] * A % P * fast(2, n - i) % P);
  }
  printf("%d\n", ans);
  return 0;
}
