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;
int dp[5001][5001];
int quickpow(int x, int y) {
  long long res = 1, base = x;
  while (y) {
    if (y & 1) res *= base, res %= 1000000007;
    y >>= 1;
    base *= base, base %= 1000000007;
  }
  return res;
}
int get_ans(int a, int b, int tot) {
  if (dp[a][b]) return dp[a][b];
  int c = tot - b;
  if (a == 0) {
    return dp[a][b] = quickpow(2, c);
  }
  return dp[a][b] = (((b ? (1ll * b * get_ans(a - 1, b, tot)) : 0ll)) +
                     ((c ? (1ll * c * get_ans(a - 1, b + 1, tot)) : 0ll))) %
                    1000000007;
}
signed main() {
  int n, k;
  scanf("%d%d", &n, &k);
  printf("%d\n", get_ans(k, 0, n));
  return 0;
}
