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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 5003, M = 1000000007;
int n, k;
int dp[N][N];
int ast(int x, int n) {
  int ans = 1;
  while (n) {
    if (n % 2 == 1) ans = (ans * 1LL * x) % M;
    x = (x * 1LL * x) % M;
    n /= 2;
  }
  return ans;
}
void solv() {
  scanf("%d%d", &n, &k);
  dp[0][0] = 1;
  for (int i = 0; i <= min(n, k); ++i) {
    for (int j = 0; j <= k; ++j) {
      dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j] * 1LL * (n - i)) % M;
      dp[i][j + 1] = (dp[i][j + 1] + dp[i][j] * 1LL * i) % M;
    }
  }
  int ans = 0;
  for (int i = 1; i <= min(n, k); ++i) {
    ans = (ans + dp[i][k] * 1LL * ast(2, n - i)) % M;
  }
  printf("%d\n", ans);
}
int main() {
  solv();
  return 0;
}
