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;
long long dp[5005][5005];
long long GET(int h) {
  long long res = 1;
  long long cur = 2;
  for (int i = 0; i <= 31; ++i) {
    if ((h >> i) & 1) res = res * cur % 1000000007;
    cur = cur * cur % 1000000007;
  }
  return res;
}
long long df(int diff, int a) {
  if (dp[diff][a]) return dp[diff][a];
  int b = n - a;
  if (!diff) return dp[diff][a] = GET(b);
  return dp[diff][a] =
             ((a ? 1ll * a * df(diff - 1, a) % 1000000007 : 0ll) +
              (b ? 1ll * b * df(diff - 1, a + 1) % 1000000007 : 0ll)) %
             1000000007;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  cout << df(k, 0);
}
