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 Mod = 1e9 + 7;
int n, k;
long long dp[5012][5012];
void Add(long long &a, long long b) { a = (a + b) % Mod; }
long long Pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % Mod;
    a = a * a % Mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> k;
  dp[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j] * j % Mod) % Mod;
    }
  }
  long long ret = 0, fac = 1;
  for (int i = 0; i <= min(n, k); i++) {
    Add(ret, fac * Pow(2, n - i) % Mod * dp[k][i] % Mod);
    fac = fac * (n - i) % Mod;
  }
  cout << ret << endl;
  return 0;
}
