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;
long long f[5005][5005];
long long n;
long long k;
long long ksm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long dfs(long long a, long long b, long long c) {
  if (f[a][b] != -1) return f[a][b];
  if (a == 0)
    f[a][b] = ksm(2, c);
  else {
    f[a][b] = 0;
    if (b != 0) f[a][b] = (f[a][b] + b * dfs(a - 1, b, c)) % 1000000007;
    if (c != 0) f[a][b] = (f[a][b] + c * dfs(a - 1, b + 1, c - 1)) % 1000000007;
  }
  return f[a][b];
}
int main() {
  cin >> n >> k;
  memset(f, -1, sizeof(f));
  cout << dfs(k, 0, n) << endl;
}
