Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long N, k;
const long long mod = 1e9 + 7;
const int maxn = 5007;
long long dp[maxn][maxn], sum[maxn];
long long mod_pow(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ans;
}
int main() {
  cin >> N >> k;
  if (N == 1) {
    return 0 * printf("1\n");
  }
  long long pre = N;
  for (int t = 0; t < min(N, 5005ll); ++t) {
    dp[0][t] = pre * mod_pow(2, N - 1 - t) % mod;
    pre = pre * (N - 1 - t) % mod;
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j <= k; ++j) {
      dp[i][j] = ((j + 1) * dp[i - 1][j] + dp[i - 1][j + 1]) % mod;
    }
  }
  printf("%lld\n", dp[k - 1][0]);
  return 0;
}