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 md = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}
int mul(int a, int b) { return ((long long)a * b) % md; }
int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  int n, k;
  cin >> n >> k;
  vector<vector<int>> dp(k + 1, vector<int>(k + 1));
  dp[0][0] = 1;
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j <= min(n, k); ++j) {
      if (j + 1 <= min(n, k)) add(dp[i + 1][j + 1], mul(dp[i][j], n - j));
      add(dp[i + 1][j], mul(dp[i][j], j));
    }
  }
  int ans = 0;
  for (int j = 1; j <= min(n, k); ++j) {
    add(ans, mul(dp[k][j], power(2, n - j)));
  }
  cout << ans << '\n';
}