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 N = 5010;
const int MOD = 1e9 + 7;
int n, k;
int dp[N][N];
int t[N];
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void mult(int &a, int b) { a = (1ll * a * b) % MOD; }
int product(int a, int b) { return (1ll * a * b) % MOD; }
int power(int b, int p) {
  int res = 1;
  while (p) {
    if (p & 1) mult(res, b);
    mult(b, b);
    p /= 2;
  }
  return res;
}
int main() {
  cin >> n >> k;
  t[0] = 1;
  for (int i = 1; i <= k; i++) t[i] = product(t[i - 1], 2);
  dp[0][0] = 1;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= i; j++) {
      if (dp[i][j] == 0) continue;
      add(dp[i + 1][j], product(j, dp[i][j]));
      add(dp[i + 1][j + 1], product(max(0, n - j), dp[i][j]));
    }
  }
  int ans = 0;
  for (int j = 0; j <= min(n, k); j++) {
    add(ans, product(dp[k][j], power(2, n - j)));
  }
  cout << ans;
  return 0;
}
