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;
long long n, k, ll;
long long mod = 1e9 + 7, qq;
long long s[5100][5100], ans;
long long ksm(long long a, long long q) {
  long long ans = 1, b = q, v = a;
  while (b) {
    if (b & 1) {
      ans *= v;
      ans %= mod;
    }
    v *= v;
    v %= mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  s[0][0] = 1;
  for (int x = 1; x <= 5000; x++) {
    for (int y = 1; y <= x; y++) {
      s[x][y] = (s[x - 1][y - 1] % mod + s[x - 1][y] % mod * y % mod) % mod;
    }
  }
  cin >> n >> k;
  qq = ksm(2, n);
  ll = 1;
  long long fan = ksm(2, mod - 2);
  for (int x = 0; x <= k; x++) {
    ans = (ans + 1ll * ll * s[k][x] % mod * qq % mod) % mod;
    ll = ll * (n - x) % mod;
    qq = 1ll * qq * fan % mod;
  }
  if (!k) {
    ans--;
  }
  cout << ans << endl;
}
