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 mod = 1e9 + 7;
int qpw(int a, int b) {
  int res = 1;
  ;
  while (b) {
    if (b & 1) res = (long long)res * a % mod;
    a = (long long)a * a % mod, b /= 2;
  }
  return res;
}
int S[5005][5005], fac[5005], dwpw[5005];
int C(int j) { return (long long)dwpw[j] * qpw(fac[j], mod - 2) % mod; }
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  S[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= k; j++)
      S[i][j] = S[i - 1][j - 1] + (long long)j * S[i - 1][j] % mod,
      S[i][j] %= mod;
  }
  fac[0] = 1;
  for (int i = 1; i <= k; i++) fac[i] = (long long)fac[i - 1] * i % mod;
  dwpw[0] = 1;
  for (int i = 1; i <= k; i++)
    dwpw[i] = (long long)dwpw[i - 1] * max(n - i + 1, 0) % mod;
  int ans = 0;
  for (int j = 0; j <= k; j++) {
    ans += (long long)S[k][j] * fac[j] % mod * C(j) % mod * qpw(2, n - j) % mod,
        ans %= mod;
  }
  printf("%d\n", ans);
  return 0;
}
