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 n, k;
const int MOD = 1e9 + 7;
int dp[5010][5010];
long long mod_pow(long long x, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * x % MOD;
    x = x * x % MOD;
    n >>= 1;
  }
  return ret;
}
int solve(int m, int k) {
  if (dp[n - m][k]) return dp[n - m][k];
  if (m == 1) return 1;
  if (k == 0) return dp[n - m][k] = mod_pow(2, m) - 1;
  long long ret = (solve(m, k - 1) - solve(m - 1, k - 1)) % MOD;
  ret = ((long long)(ret) * (m) % MOD);
  if (ret < 0) ret += MOD;
  return dp[n - m][k] = ret;
}
int main() {
  scanf("%d%d", &(n), &(k));
  cout << solve(n, k);
}
