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 MAX = 5005;
long long n, k, mod = 1e9 + 7, a[MAX], c[2][MAX];
long long Pow(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long res = Pow(a, n / 2);
  res = res * res % mod;
  if (n % 2) res = res * a % mod;
  return res;
}
int main() {
  cin >> n >> k;
  a[0] = 1;
  for (int i = 1; i <= k; i++) a[i] = a[i - 1] * (n - i + 1) % mod;
  for (int i = 0; i <= k; i++) a[i] = a[i] * Pow(2, n - i) % mod;
  c[0][0] = 1;
  int id = 0;
  for (int i = 1; i <= k; i++) {
    id ^= 1;
    for (int j = 0; j <= k; j++) c[id][j] = 0;
    for (int j = 0; j <= k; j++) {
      c[id][j + 1] = (c[id][j + 1] + c[id ^ 1][j]) % mod;
      c[id][j] = (c[id][j] + j * c[id ^ 1][j]) % mod;
    }
  }
  long long res = 0;
  for (int i = 0; i <= k; i++) res = (res + a[i] * c[id][i]) % mod;
  cout << res;
}
