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;
int dp[5005][5005];
int exp(int base, int r) {
  if (r == 0) return 1;
  int ans = exp(base, r / 2);
  if (r % 2 == 0)
    return (long long int)ans * ans % 1000000007;
  else
    return (long long int)base * ans % 1000000007 * ans % 1000000007;
}
int go(int dif, int b, int tot) {
  if (dp[dif][b] >= 0) return dp[dif][b];
  int c = tot - b;
  if (dif == 0) {
    return dp[dif][b] = exp(2, c);
  }
  return dp[dif][b] =
             ((b ? 1LL * b * go(dif - 1, b, tot) : 0LL) % 1000000007 +
              (c ? 1LL * c * go(dif - 1, b + 1, tot) : 0LL) % 1000000007) %
             1000000007;
}
int main() {
  int n, k;
  memset(dp, -1, sizeof(dp));
  scanf(" %d%d", &n, &k);
  printf("%d\n", go(k, 0, n));
  return 0;
}
