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 MOD = 1e9 + 7;
long long st[2][5010];
long long rp[5010];
long long n, k;
long long tp(long long n) {
  if (!n) return 1;
  long long t = tp(n / 2);
  return t * t % MOD * (n % 2 ? 2 : 1);
}
int main() {
  scanf("%lld%lld", &n, &k);
  int i, j;
  st[0][0] = 1;
  for (i = 0; i < k; i++) {
    for (j = 1; j <= i + 1; j++)
      st[1][j] = (st[0][j] * j % MOD + st[0][j - 1]) % MOD;
    for (j = 0; j <= i + 1; j++) st[0][j] = st[1][j];
  }
  rp[0] = n;
  for (i = 1; i <= k; i++) rp[i] = rp[i - 1] * (n - i) % MOD;
  long long res = 0;
  for (i = 0; i < k; i++) {
    res += st[0][i + 1] * tp(n - 1 - i) % MOD * rp[i] % MOD;
    res %= MOD;
  }
  printf("%lld\n", res);
  return 0;
}
