Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int K = 5e3 + 5;
const int mod = 1e9 + 7;
long long s2[K][K];
int n, k;
long long fp(long long a, long long b) {
  if (!b) return 1ll;
  long long t = fp(a, b / 2);
  if (b & 1)
    return t * t % mod * a % mod;
  else
    return t * t % mod;
}
signed main() {
  cin >> n >> k;
  s2[0][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++)
      s2[i][j] = (s2[i - 1][j - 1] + j * s2[i - 1][j]) % mod;
  long long pw2 = fp(2, n);
  long long inv2 = fp(2, mod - 2);
  long long downp = 1ll;
  long long ans = 0ll;
  for (int i = 0; i <= k; i++) {
    (ans += downp * s2[k][i] % mod * pw2 % mod) %= mod;
    (downp *= (n - i)) %= mod;
    (pw2 *= inv2) %= mod;
  }
  if (!k) --ans;
  cout << ans << endl;
}