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 = 1e9 + 7, INV2 = 5e8 + 4;
int S[5007][5007], N, K;
int poww(long long a, int b) {
  int times = 1;
  while (b) {
    if (b & 1) times = times * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return times;
}
int main() {
  cin >> N >> K;
  S[1][1] = 1;
  for (int i = 2; i <= K; ++i)
    for (int j = 1; j <= K; ++j)
      S[i][j] = (S[i - 1][j - 1] + 1ll * j * S[i - 1][j]) % MOD;
  long long fac = 1, binom = 1, pow2 = poww(2, N), ans = 0;
  for (int i = 1; i <= K; ++i) {
    binom = binom * (N - i + 1) % MOD * poww(i, MOD - 2) % MOD;
    pow2 = pow2 * INV2 % MOD;
    fac = fac * i % MOD;
    ans = (ans + S[K][i] * fac % MOD * binom % MOD * pow2 % MOD) % MOD;
  }
  cout << ans;
  return 0;
}
