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 = 5000 + 10;
const int mod = 1000000000 + 7;
long long S[maxn][maxn];
long long qpow(long long a, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) {
      res = res * a % mod;
    }
    a = a * a % mod;
    k >>= 1;
  }
  return res;
}
long long inv(long long x) { return qpow(x, mod - 2); }
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  if (n <= k) {
    long long ans = 0, c = 1;
    for (int i = 1; i <= n; i++) {
      c = c * (n - i + 1) % mod * inv(i) % mod;
      ans = (ans + c * qpow(i, k) % mod) % mod;
    }
    printf("%lld\n", ans);
  } else {
    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] + j * S[i - 1][j]) % mod;
      }
    }
    long long ans = 0;
    long long t1 = 1, t2 = 1;
    for (int j = 1; j <= k; j++) {
      t1 = t1 * j % mod;
      t2 = t2 * (n - j + 1) % mod * inv(j) % mod;
      ans = (ans + S[k][j] * t1 % mod * t2 % mod * qpow(2, n - j) % mod) % mod;
    }
    printf("%lld\n", ans);
  }
  getchar();
  getchar();
  return 0;
}
