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 n, k, i, j, a[5010][5010];
long long first(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long t = first(a, b / 2);
    return (t * t) % 1000000007;
  }
  return (first(a, b - 1) * a) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cin >> n >> k;
  for (i = 0; i <= k; i++) {
    if (i + n > k)
      a[0][i] = (first(2, i + n - k) + 1000000007 - 1) % 1000000007;
  }
  for (i = 1; i <= k; i++) {
    for (j = k; j >= i && j + n > k; j--) {
      a[i][j] = ((j + n - k) *
                 ((a[i - 1][j] - a[i - 1][j - 1] + 1000000007) % 1000000007)) %
                1000000007;
    }
  }
  cout << a[k][k] << endl;
  return 0;
}
