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 inf = 0x3f3f3f3f;
const long long mod = 1000000007;
long long S[5050][5050];
long long m, n, x, y, res;
long long quick(long long x, long long k, long long mod) {
  long long res = 1;
  while (k > 0) {
    if (k & 1) res = res * x % mod;
    x = x * x % mod;
    k >>= 1;
  }
  return res;
}
void init() {
  for (long long i = 1; i <= n + 2; i++) {
    for (long long j = 1; j <= i; j++) {
      S[i][j] = (S[i - 1][j - 1] + j * S[i - 1][j] % mod) % mod;
    }
  }
}
int main() {
  cin >> m >> n;
  S[0][0] = 1;
  init();
  long long k = 1;
  for (long long j = 0; j <= n; j++) {
    res = (res + S[n][j] * quick(2, m - j, mod) % mod * k % mod) % mod;
    k = k * (m - j) % mod;
  }
  cout << (res + mod) % mod << endl;
}
