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;
const int maxm = 5007;
long long dp[maxm][maxm];
int qp(int n, int k) {
  if (k < 0) return 0;
  long long ans = 1, u = n;
  while (k) {
    if (k & 1) {
      ans = (ans * u) % mod;
    }
    u = (u * u) % mod;
    k >>= 1;
  }
  return ans;
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i <= k; ++i) {
    dp[i][0] = qp(2, n - i);
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j <= k - i; ++j) {
      dp[j][i] = ((n - j) * dp[j + 1][i - 1] + j * dp[j][i - 1]) % mod;
    }
  }
  cout << dp[0][k] << endl;
  return 0;
}
