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;
int n, k, ans, s[5005][5005];
inline int read() {
  int num = 0;
  char cc = getchar();
  while (cc < 48 || cc > 57) cc = getchar();
  while (cc > 47 && cc < 58)
    num = (num << 1) + (num << 3) + (cc ^ 48), cc = getchar();
  return num;
}
inline int power(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  n = read(), k = read();
  s[0][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++)
      s[i][j] = (s[i - 1][j - 1] + 1ll * s[i - 1][j] * j) % mod;
  int fac = 1, mul2 = power(2, n), inv = power(2, mod - 2);
  for (int i = 0; i <= k; i++) {
    (ans += 1ll * s[k][i] * fac % mod * mul2 % mod) %= mod;
    fac = 1ll * fac * (n - i) % mod;
    mul2 = 1ll * mul2 * inv % mod;
  }
  if (!k) ans--;
  printf("%d", ans);
}
