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 N = 5005;
const long long p = 1e9 + 7;
long long dp[N], dp1[N];
long long modexp(long long a, long long b, long long p) {
  long long ans = 1LL;
  a %= p;
  while (b) {
    if (b & 1) ans = ans * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return ans;
}
int main() {
  memset(dp, 0, sizeof(dp));
  int n, k;
  scanf("%d%d", &n, &k);
  dp[0] = 1LL;
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < N; j++) {
      if (j) {
        dp1[j] += j * 1LL * dp[j];
        dp1[j] %= p;
      }
      if (dp[j]) {
        dp1[j + 1] += (n - j) * 1LL * dp[j];
        dp1[j + 1] %= p;
      }
    }
    for (int j = 0; j < N; j++) dp[j] = dp1[j];
    memset(dp1, 0, sizeof(dp1));
  }
  long long ans = 0;
  for (int i = 0; i <= min(n, N - 1); i++)
    ans = (ans + dp[i] * modexp(2, n - i, p) % p) % p;
  printf("%lld\n", ans);
  return 0;
}
