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 long long int mod = 1e9 + 7;
long long int c[2][5010];
long long int fxp(long long int b, long long int x) {
  long long int a = 1;
  for (; x != 0; b = b * b % mod, x >>= 1)
    if (x & 1ll) a = a * b % mod;
  return a;
}
int main() {
  int n, k;
  cin >> n >> k;
  int rw = 0;
  c[rw][0] = 1;
  for (int i = 0; i < k; i++) {
    rw ^= 1;
    for (long long int j = 0; j <= i; j++) c[rw][j] = 0;
    for (long long int j = 0; j <= i; j++) {
      long long int a = n - j;
      c[rw][j] = (c[rw][j] + c[rw ^ 1][j] * j) % mod;
      c[rw][j + 1] = (c[rw][j + 1] + c[rw ^ 1][j] * a) % mod;
    }
  }
  long long int ans = 0;
  for (int j = 0; j <= k && n - j >= 0; j++) {
    int a = n - j;
    ans += c[rw][j] * fxp(2, a) % mod;
  }
  printf("%lld\n", ans % mod);
  return 0;
}
