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 N = 5e3 + 5, mod = 1e9 + 7;
int ans, n, k, dp[N][N], tmp;
int zrb(int a, int b) { return ((a * 1LL * b % mod + 2LL * mod) % mod); }
int jm(int a, int b) { return ((a + b + 2LL * mod) % mod); }
int pw(int a, int b) {
  if (!b) return (1LL);
  int tmp = pw(a, b / 2);
  return (zrb(zrb(tmp, tmp), b & 1 ? a : 1LL));
}
int main() {
  scanf("%d %d", &n, &k);
  dp[0][0] = 1;
  for (int i = 1; i <= k + 1; i++)
    for (int j = 1; j <= i; j++) {
      dp[i][j] = jm(zrb(j, dp[i - 1][j]), zrb(n - j + 1, dp[i - 1][j - 1]));
    }
  for (int i = 1; i <= min(k, n); i++)
    ans = jm(ans, zrb(dp[k][i], pw(2, n - i)));
  printf("%d", ans);
}
