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;
long long st[5010][5010], n, k, ans;
long long ksm(long long aa, long long bb) {
  long long res = 1;
  while (bb > 0) {
    if (bb & 1) res = res * aa % mod;
    bb >>= 1, aa = aa * aa % mod;
  }
  return res;
}
int main() {
  scanf("%lld%lld", &n, &k);
  st[1][1] = 1;
  for (int i = 2; i <= k; i++)
    for (int j = 1; j <= i; j++)
      st[i][j] = (st[i - 1][j] * j + st[i - 1][j - 1]) % mod;
  for (int i = 0; i <= k; i++) {
    long long res = 1;
    for (int j = 1; j <= i; j++) res = res * (n - j + 1) % mod;
    res = res * ksm(2, n - i) % mod;
    res = res * st[k][i];
    ans = (ans + res) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
