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 N = 5005, MOD = 1e9 + 7;
int dp[N][N];
void upd(int &x, int y) { (x += y) %= MOD; }
int mpow(int x, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1LL * res * x % MOD;
    n >>= 1;
    x = 1LL * x * x % MOD;
  }
  return res;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  dp[0][0] = 1;
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j <= i; ++j)
      if (dp[i][j]) {
        upd(dp[i + 1][j], 1LL * j * dp[i][j] % MOD);
        upd(dp[i + 1][j + 1], 1LL * (n - j) * dp[i][j] % MOD);
      }
  }
  int ans = 0;
  for (int j = 1; j <= min(n, k); ++j) {
    upd(ans, 1LL * mpow(2, n - j) * dp[k][j] % MOD);
  }
  printf("%d\n", ans);
}
