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 maxn = 5010, mod = 1e9 + 7;
int _pow(int x, int n) {
  int ret = 1;
  for (; n; n >>= 1, x = 1LL * x * x % mod)
    if (n & 1) ret = 1LL * ret * x % mod;
  return ret;
}
int s[maxn][maxn], fac[maxn], ifac[maxn], n, K;
int main() {
  scanf("%d%d", &n, &K);
  s[0][0] = fac[0] = 1;
  for (register int i = 1; i <= K; ++i) {
    for (register int k = 1; k <= i; ++k) {
      s[i][k] = (s[i - 1][k - 1] + 1LL * s[i - 1][k] * k % mod) % mod;
    }
  }
  int ret = 0, now = _pow(2, n), inv2 = _pow(2, mod - 2), val = 1;
  for (register int i = 1; i <= min(K, n); ++i) {
    now = 1LL * now * inv2 % mod;
    val = 1LL * val * (n - i + 1) % mod;
    ret = (ret + 1LL * s[K][i] * val % mod * now % mod) % mod;
  }
  printf("%d\n", ret);
  return 0;
}
