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 s[5010][5010], ans;
int n, k;
void predo() {
  s[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      s[i][j] = (s[i - 1][j - 1] + s[i - 1][j] * j) % mod;
    }
  }
}
long long qp(long long x, int t) {
  long long res = 1;
  for (; t; t >>= 1, x = x * x % mod) {
    if (t & 1) res = res * x % mod;
  }
  return res;
}
int main() {
  scanf("%d %d", &n, &k);
  predo();
  long long las = 1;
  for (int j = 0; j <= min(n, k); j++) {
    ans += s[k][j] * las % mod * qp(2, n - j) % mod;
    ans %= mod;
    las = las * (n - j) % mod;
  }
  if (n == 0) ans = 0;
  printf("%lld", ans % mod);
  return 0;
}
