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, dp[2][5005], cur, ans;
long long pow(long long x, long long y) {
  long long res = 1, a = x;
  for (; y; y >>= 1) {
    if (y & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
int main() {
  cin >> n >> k;
  dp[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    cur = !cur;
    for (int j = 1; j <= i; j++)
      dp[cur][j] =
          (dp[!cur][j] * j + dp[!cur][j - 1] * (n - j + 1)) % 1000000007;
    dp[0][0] = 0;
  }
  for (int i = 1; i <= min(n, k); i++)
    ans = (ans + dp[cur][i] * pow(2ll, n - i) % 1000000007) % 1000000007;
  printf("%I64d", ans);
}
