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>
int64_t dp[5005][5005];
int64_t pwr(int64_t x, int p) {
  if (p == 0) return 1LL;
  if (p & 1) return x * pwr(x, p - 1) % 1000000007LL;
  return pwr(x * x % 1000000007LL, p >> 1);
}
int64_t foo(int s, int p, int q) {
  if (dp[s][p] != -1) return dp[s][p];
  int64_t res;
  if (s == 0)
    res = pwr(2, q);
  else if (q == 0)
    res = pwr(p, s);
  else
    res = p * foo(s - 1, p, q) + q * foo(s - 1, p + 1, q - 1);
  return dp[s][p] = res % 1000000007LL;
}
int main(void) {
  int n, k;
  scanf("%d%d", &n, &k);
  memset(dp, -1, sizeof(dp));
  printf("%d\n", foo(k, 0, n));
  return 0;
}
