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 long long mod = 1e9 + 7;
long long int mypow(long long int a, int b) {
  if (b < 0) return 0;
  long long int ans = 1;
  for (; b; b >>= 1, a = a * a % mod) {
    if (b & 1) ans = ans * a % mod;
  }
  return ans;
}
void add(long long int &a, long long int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int main() {
  int n, k;
  cin >> n >> k;
  long long int dp[k + 1][k + 1];
  for (int i = (0); i < (k + 1); i++) {
    for (int j = (0); j < (k + 1); j++) {
      if (i) {
        dp[i][j] = 0;
        add(dp[i][j], j * dp[i - 1][j] % mod);
        if (n > j) add(dp[i][j], (n - j) * dp[i - 1][j + 1] % mod);
      } else {
        dp[i][j] = mypow(2, n - j);
      }
    }
  }
  cout << dp[k][0] << endl;
}