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 = 5555;
const int mod = 1e9 + 7;
int dp[N][N];
int pw2[N];
int n, K;
int bm(int b, int p) {
  if (p == 0) return 1;
  int t = bm(b, p >> 1);
  t = 1ll * t * t % mod;
  if (p & 1) return 1ll * t * b % mod;
  return t;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> K;
  dp[1][1] = n;
  for (int i = 1; i <= K; i++) {
    for (int j = 1; j <= min(i, n); j++) {
      dp[i + 1][j + 1] += 1ll * dp[i][j] * (n - j) % mod;
      dp[i + 1][j + 1] %= mod;
      dp[i + 1][j] += 1ll * dp[i][j] * j % mod;
      dp[i + 1][j] %= mod;
    }
  }
  int ways = 0;
  for (int j = 1; j <= min(K, n); j++) {
    ways += 1ll * dp[K][j] * bm(2, n - j) % mod;
    ways %= mod;
  }
  cout << ways << endl;
  return 0;
}
