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 long long mod = 1e9 + 7;
long long n, k, ans, dp[5500][5500];
long long quick(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}
int main() {
  ans = 0;
  scanf("%lld%lld", &n, &k);
  dp[1][1] = n;
  for (int i = 1; i < k; i++) {
    for (int j = 1; j <= i; j++) {
      dp[i + 1][j] += dp[i][j] * j;
      dp[i + 1][j] %= mod;
      dp[i + 1][j + 1] += dp[i][j] * (n - j);
      dp[i + 1][j + 1] %= mod;
    }
  }
  for (int i = 1; i <= k; i++) {
    ans = (ans + quick(2, n - i) * dp[k][i] % mod) % mod;
  }
  printf("%lld\n", ans % mod);
  return 0;
}
