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 = 1000000007;
const int inv2 = mod / 2 + 1;
int n, k, S[5010];
int pow(long long a, int b, int m) {
  long long t = 1;
  for (; b; b >>= 1, a = a * a % m)
    if (b & 1) t = t * a % m;
  return t;
}
void CalS() {
  S[1] = 1;
  for (int i = 2; i <= k; i++)
    for (int j = i; j; j--) S[j] = (1LL * S[j] * j + S[j - 1]) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  CalS();
  int P = 1, M = pow(2, n, mod), ans = 0;
  for (int i = 1; i <= k; i++) {
    P = 1LL * P * (n - i + 1) % mod;
    M = 1LL * M * inv2 % mod;
    ans = (1LL * ans + 1LL * S[i] * P % mod * (long long)M % mod) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
