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 sz = 5e3 + 10, mod = 1e9 + 7;
int dp[sz][sz];
int po(int x, int y) {
  int re = 1;
  while (y) {
    if (y & 1) re = (1ll * re * x) % mod;
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return re;
}
int main() {
  int n, k, an = 0;
  cin >> n >> k;
  dp[0][0] = 1;
  for (int a = 0; a < k; a++) {
    for (int b = 0; b <= min(a, n); b++) {
      dp[a + 1][b] = (dp[a + 1][b] + 1ll * b * dp[a][b]) % mod;
      if (b + 1 <= n)
        dp[a + 1][b + 1] = (dp[a + 1][b + 1] + 1ll * (n - b) * dp[a][b]) % mod;
    }
  }
  for (int a = 0; a <= min(k, n); a++) {
    an = (an + 1ll * dp[k][a] * po(2, n - a)) % mod;
  }
  cout << an;
}
