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 int mod = 1e9 + 7, MAXN = 5050;
int qpow(int x, int y) {
  if (y < 0) return qpow(qpow(x, -y), mod - 2);
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1ll * ret * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return ret;
}
int s2[MAXN][MAXN], rua[MAXN], p2[MAXN];
int n, t;
int main() {
  scanf("%d%d", &n, &t);
  int i, j;
  rua[0] = 1;
  s2[0][0] = 1;
  p2[0] = qpow(2, n);
  for (i = 1; i <= t; i++)
    rua[i] = 1ll * rua[i - 1] * (n - i + 1) % mod, p2[i] = qpow(2, n - i);
  for (i = 1; i <= t; i++)
    for (j = 1; j <= i; j++)
      s2[i][j] = (1ll * j * s2[i - 1][j] % mod + s2[i - 1][j - 1]) % mod;
  int ans = 0;
  for (i = 0; i <= t; i++)
    ans = (ans + 1ll * s2[t][i] * rua[i] % mod * p2[i] % mod) % mod;
  printf("%d\n", (ans + mod) % mod);
  return 0;
}
