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;
int c[5050][5050];
int mod = 1e9 + 7;
int C(int n, int k) {
  if (n == k || k == 0) return c[n][k] = 1;
  if (c[n][k]) return c[n][k];
  return c[n][k] = (C(n - 1, k - 1) + C(n - 1, k)) % mod;
}
long long bp(long long a, int n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    n >>= 1;
  }
  return res;
}
int f[5050][5050];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  if (n <= k) {
    long long res = 0;
    for (int i = 1; i <= n; i++) {
      res += ((long long)bp(i, k) * C(n, i)) % mod;
      res %= mod;
    }
    cout << res;
  } else {
    f[1][0] = n;
    long long res = 0;
    for (int i = 2; i <= k; i++) {
      for (int j = 0; j <= i; j++) {
        f[i][j] = ((long long)f[i - 1][j] * (n - j)) % mod;
        if (j) f[i][j] -= ((long long)(n - j) * f[i - 1][j - 1]) % mod;
        f[i][j] = (f[i][j] % mod + mod) % mod;
      }
    }
    for (int i = 0; i <= k; i++) {
      res += ((long long)f[k][i] * bp(2, n - i - 1)) % mod;
      res %= mod;
    }
    cout << res;
  }
  return 0;
}
