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;
long long power(long long a, long long b, long long m) {
  long long x = 1 % m;
  a %= m;
  while (b) {
    if ((1 & b)) x = x * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return x;
}
int naive_stirling[5005][5005];
void prepare_stirling() {
  memset(naive_stirling, 0, sizeof(naive_stirling));
  naive_stirling[0][0] = 1;
  for (int n = 1; n < 5005; n++) {
    for (int k = 1; k <= n; k++) {
      naive_stirling[n][k] =
          (k * 1LL * naive_stirling[n - 1][k] + naive_stirling[n - 1][k - 1]) %
          1000000007;
    }
  }
}
int main() {
  prepare_stirling();
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  int it = min(n, k);
  long long ans = 0;
  long long comb = 1;
  long long fact = 1;
  for (int i = 0; i <= it; i++) {
    long long tmp = comb * naive_stirling[k][i];
    tmp %= 1000000007;
    tmp *= fact;
    tmp %= 1000000007;
    tmp = tmp * power(2, n - i, 1000000007) % 1000000007;
    ans += tmp;
    comb = comb * (n - i) % 1000000007;
    comb = comb * power(i + 1, 1000000007 - 2, 1000000007) % 1000000007;
    fact = fact * (i + 1) % 1000000007;
  }
  ans %= 1000000007;
  printf("%lld\n", ans);
}
