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;
const int N = 5e3 + 10, mod = 1e9 + 7;
int dp[N][N];
int pw(int x, int y) {
  int ret = 1;
  for (; y; y /= 2) {
    if (y & 1) ret = 1ll * ret * x % mod;
    x = 1ll * x * x % mod;
  }
  return ret;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int ans = 0;
  dp[0][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) {
      dp[i][j] = (1ll * dp[i - 1][j] * j % mod +
                  1ll * dp[i - 1][j - 1] * (n - j + 1) % mod) %
                 mod;
      if (i == k) (ans += 1ll * dp[i][j] * pw(2, n - j) % mod) %= mod;
    }
  cout << ans;
  return 0;
}