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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
const int N = 1e5 + 7;
const int M = 1e6 + 7;
const long long mod = 1e9 + 7;
long long n, k;
long long st[5001][5001];
long long power(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
void init() {
  st[1][1] = 1;
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      st[i][j] =
          (((long long)j * st[i - 1][j]) % mod + (st[i - 1][j - 1])) % mod;
    }
  }
}
long long A(long long n, long long m) {
  long long ans = 1;
  while (m--) {
    ans *= n--;
    ans %= mod;
  }
  return ans;
}
int main() {
  cin >> n >> k;
  init();
  long long mmin = min(n, k);
  long long ans = 0;
  for (int i = 0; i <= mmin; i++) {
    ans += ((st[k][i]) * A(n, i) % mod * power(2, n - i) % mod) % mod;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}
