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 K = 5e3 + 10, mod = 1e9 + 7;
bool seen[K][K];
long long dp[K][K];
long long solve(int a, long long b, int c);
long long pwr(long long x, long long e);
int main() {
  long long n, k;
  cin >> n >> k;
  cout << solve(0, n, k);
}
long long solve(int a, long long b, int c) {
  if (seen[a][c]) {
    return dp[a][c];
  } else {
    long long ans = 0;
    if (b < 0) {
      ans = 0;
    } else if (!c) {
      ans = pwr(2, b);
    } else {
      ans = b * solve(a + 1, b - 1, c - 1);
      ans += a * solve(a, b, c - 1);
      ans %= mod;
    }
    seen[a][c] = true;
    dp[a][c] = ans;
    return ans;
  }
}
long long pwr(long long x, long long e) {
  long long res = 1;
  while (e) {
    if (e & 1) {
      res = (res * x) % mod;
    }
    x = (x * x) % mod;
    e >>= 1;
  }
  return res;
}
