Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n, k;
long long f[5005][5005];
int qucp(long long x, int y) {
  long long an = 1;
  while (y) {
    if (y & 1) an = an * x % mod;
    y /= 2;
    x = x * x % mod;
  }
  return an;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  f[1][1] = n;
  int i, j;
  long long ans = 0;
  for (i = 1; i < k; ++i)
    for (j = 1; j <= i; ++j) {
      (f[i + 1][j] += f[i][j] * j) %= mod;
      (f[i + 1][j + 1] += f[i][j] * (n - j)) %= mod;
    }
  for (i = 1; i <= k; ++i) (ans += f[k][i] * qucp(2, n - i)) %= mod;
  cout << ans;
  return 0;
}