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;
long long dp[5001][5001];
int n, k;
long long power(int num) {
  if (num == 0) return 1;
  if (num == 1) return 2;
  long long ans = power(num / 2);
  ans *= ans;
  ans %= 1000000007;
  if (num % 2) {
    ans *= 2;
    ans %= 1000000007;
  }
  return ans;
}
long long fun(int pos, int k) {
  if (dp[pos][k] != -1000000007) return dp[pos][k];
  if (k == 0) return power(n - pos);
  long long ans = (n + 1 - pos) * fun(pos, k - 1);
  ans %= 1000000007;
  ans -= (n - pos) * fun(pos + 1, k - 1);
  ans %= 1000000007;
  dp[pos][k] = ans;
  return ans;
}
int main() {
  scanf("%d %d", &n, &k);
  int i, j;
  for (i = 0; i <= 5000; i++) {
    for (j = 0; j <= 5000; j++) {
      dp[i][j] = -1000000007;
    }
  }
  long long ans = n * fun(1, k - 1);
  ans %= 1000000007;
  ans += 1000000007;
  ans %= 1000000007;
  printf("%lld\n", ans);
  return 0;
}
