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;
long long qpow(long long b, long long p) {
  long long ret = 1LL;
  while (p) {
    if (p & 1) ret = ret * b % Mod;
    b = b * b % Mod;
    p >>= 1;
  }
  return ret;
}
long long stirl[5050][5050];
long long n, k;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  stirl[1][1] = 1;
  for (long long i = 2; i <= k; i++) {
    for (long long j = 1; j <= k; j++) {
      stirl[i][j] = (stirl[i - 1][j - 1] + j * stirl[i - 1][j]) % Mod;
    }
  }
  long long ans = 0;
  long long now_pow = 1LL;
  for (long long i = 0; i <= k && i <= n; i++) {
    ans += stirl[k][i] * now_pow % Mod * qpow(2, n - i) % Mod;
    now_pow = (now_pow * (n - i)) % Mod;
  }
  cout << ans % Mod << endl;
  return 0;
}
