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 mod = 1e9 + 7;
long long qpow(long long x, long long k) {
  return k == 0 ? 1
                : 1ll * qpow(1ll * x * x % mod, k >> 1) * (k & 1 ? x : 1) % mod;
}
const int md = 1e9 + 7;
const int Maxn = 5005;
int n, k;
long long dp[Maxn][Maxn];
int main() {
  scanf("%d%d", &n, &k);
  dp[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      dp[i][j] = (dp[i - 1][j] * j + dp[i - 1][j - 1] * (n - j + 1)) % md;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= min(n, k); i++) {
    ans = (ans + dp[k][i] * qpow(2, n - i)) % md;
  }
  printf("%lld\n", ans);
  return 0;
}
