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 maxn = 5e3 + 9, mod = 1e9 + 7, inv2 = 500000004;
inline int Pow(int base, int b) {
  int ret(1);
  while (b) {
    if (b & 1) ret = 1ll * ret * base % mod;
    base = 1ll * base * base % mod;
    b >>= 1;
  }
  return ret;
}
int ans[maxn][maxn];
inline void Fir(int n) {
  ans[1][1] = 1;
  for (int i = 2; i <= n; ++i)
    for (int j = 1; j <= i; ++j)
      ans[i][j] =
          1ll * (ans[i - 1][j - 1] + 1ll * j * ans[i - 1][j] % mod) % mod;
}
inline int Get(int l, int r) {
  int ret(1);
  for (int i = l; i <= r; ++i) ret = 1ll * ret * i % mod;
  return ret;
}
int n, k, ret;
int main() {
  scanf("%d%d", &n, &k);
  Fir(k);
  for (int j = 0, val1 = 1, val2 = Pow(2, n); j <= k; ++j,
           val1 = 1ll * val1 * (n - j + 1) % mod,
           val2 = 1ll * val2 * inv2 % mod)
    ret = 1ll * (ret + 1ll * ans[k][j] * val1 % mod * val2 % mod) % mod;
  printf("%d ", ret);
}
