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;
using ll = long long;
using ld = long double;
const int MOD = 1E9 + 7;
const int MAXN = 5050;
int dp[MAXN][MAXN];
ll qpow(int n, int k) {
  ll res = 1;
  if (k < 0) return 0LL;
  while (k) {
    if (k & 1) res = res * 1LL * n % MOD;
    k >>= 1;
    n = n * 1LL * n % MOD;
  }
  return res;
}
int main() {
  dp[0][0] = 1;
  int N, K;
  scanf("%d %d", &N, &K);
  for (int i = 0; i < K; i++) {
    for (int j = 0; j <= i; j++) {
      if (dp[i][j] == 0) continue;
      dp[i + 1][j] = (dp[i + 1][j] + j * 1LL * dp[i][j]) % MOD;
      dp[i + 1][j + 1] = (dp[i + 1][j + 1] + (N - j) * 1LL * dp[i][j]) % MOD;
    }
  }
  ll res = 0;
  for (int i = 0; i <= K; i++) {
    res = (res + dp[K][i] * qpow(2, N - i)) % MOD;
  }
  printf("%lld\n", res);
  return 0;
}
