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 long long mod = 1e9 + 7;
const int N = 5002;
long long S[N][N];
long long Quick_Pow(long long x, int k) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * x % mod;
    x = x * x % mod;
    k /= 2;
  }
  return ret;
}
void Pre() {
  S[0][0] = 1;
  for (int i = 1; i <= 5001; ++i) {
    for (int j = 1; j <= i; ++j) {
      S[i][j] = (S[i - 1][j - 1] + S[i - 1][j] * j) % mod;
    }
  }
}
long long F(int r, int l) {
  long long ret = 1;
  for (int i = l; i <= r; ++i) ret = ret * i % mod;
  return ret;
}
int main() {
  Pre();
  int n, k;
  cin >> n >> k;
  long long ret = 0;
  for (int i = 1; i <= k; ++i)
    ret = (ret + Quick_Pow(2, n - i) * F(n, n - i + 1) % mod * S[k][i] % mod) %
          mod;
  cout << ret;
  return 0;
}
