Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 5003, mod = 1e9 + 7;
long long n, k, S[N][N], ans;
inline long long kasumi(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (long long)res * a % mod;
    a = (long long)a * a % mod;
    b >>= 1;
  }
  return res;
}
signed main() {
  scanf("%d%d", &n, &k);
  S[0][0] = 1;
  for (register long long i = 1; i <= k; i++)
    for (register long long j = 1; j <= i; j++)
      S[i][j] = ((long long)S[i - 1][j] * j + S[i - 1][j - 1]) % mod;
  long long fac = 1, cho = 1;
  for (register long long i = 1; i <= k && i <= n; i++) {
    cho = (long long)cho * (n - i + 1) % mod * kasumi(i, mod - 2) % mod;
    fac = (long long)fac * i % mod;
    ans = (ans + (long long)fac * cho % mod * kasumi(2, n - i) % mod * S[k][i] %
                     mod) %
          mod;
  }
  printf("%d", ans);
}