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;
namespace FGF {
int n, m;
const int N = 5005, mo = 1e9 + 7;
int s[N][N], ans;
int qpow(int x, int y) {
  int s = 1;
  while (y) {
    if (y & 1) s = 1LL * s * x % mo;
    x = 1LL * x * x % mo, y >>= 1;
  }
  return s;
}
void work() {
  scanf("%d%d", &n, &m);
  s[0][0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 0; j <= m; j++)
      s[i][j] = (1LL * j * s[i - 1][j] + s[i - 1][j - 1]) % mo;
  int c = 1, t = qpow(2, n), inv2 = (mo + 1) / 2;
  for (int i = 0; i <= m; i++)
    ans = (ans + 1LL * s[m][i] * c % mo * t % mo) % mo,
    c = 1ll * c * (n - i) % mo, t = 1ll * t * inv2 % mo;
  printf("%d", ans);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}
