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 M = 1000000007;
const long long N = 5005;
long long s[N][N], fn[N], n, k, ans = 0;
long long kpw(long long x, long long y) {
  long long ans = 1LL, tmp = x;
  while (y) {
    if (y & 1) ans = (ans * tmp) % M;
    tmp = (tmp * tmp) % M;
    y >>= 1;
  }
  return ans;
}
void init_s2() {
  s[0][0] = 1;
  for (long long i = 1; i < N; i++) {
    for (long long j = 1; j <= i; j++) {
      s[i][j] = (s[i - 1][j - 1] + j * s[i - 1][j]) % M;
    }
  }
}
int main() {
  cin >> n >> k;
  init_s2();
  fn[1] = n % M;
  for (long long i = 2; i <= min(n, k); i++)
    fn[i] = (fn[i - 1] * (n - i + 1)) % M;
  for (long long i = 1; i <= min(n, k); i++)
    ans = (ans + (s[k][i] * fn[i]) % M * kpw(2LL, n - i) % M) % M;
  cout << ans << '\n';
  return 0;
}
