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>
const int MAXK = 5002;
const int MOD = 1e9 + 7;
const int INV = 500000004;
int n, K;
int S[MAXK | 1][MAXK | 1];
inline int read() {
  register int x = 0, v = 1;
  register char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') v = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * v;
}
inline int f_pow(int x, int y) {
  int res = 1, base = x;
  while (y > 0) {
    if (y & 1) res = 1LL * res * base % MOD;
    base = 1LL * base * base % MOD;
    y >>= 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] = (1LL * j * S[i - 1][j] % MOD + S[i - 1][j - 1]) % MOD;
  int ans = 0;
  for (int a = n, b = f_pow(2, n), i = 1; i <= K; ++i) {
    b = 1LL * b * INV % MOD;
    ans = (ans + 1LL * a * S[K][i] % MOD * b % MOD) % MOD;
    a = 1LL * a * (n - i) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}
