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 read() {
  char c;
  int sum = 0, f = 1;
  c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum * f;
}
int n, k;
long long dp[5005][5005];
long long ksm(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return ret;
}
long long dfs(int a, int b) {
  if (b > n) return 0;
  if (dp[a][b] >= 0) return dp[a][b];
  int c = n - b;
  if (a == 0) return dp[a][b] = ksm(2, c);
  dp[a][b] = (b * dfs(a - 1, b) + c * dfs(a - 1, b + 1)) % 1000000007;
  return dp[a][b];
}
int main() {
  n = read();
  k = read();
  memset(dp, -1, sizeof(dp));
  printf("%lld\n", dfs(k, 0));
}
