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;
long long mod = 1e9 + 7;
long long mpow(long long b, long long e) {
  if (e == 0) return 1;
  if (e % 2) return (b * mpow(b, e - 1)) % mod;
  long long tmp = mpow(b, e / 2);
  return (tmp * tmp) % mod;
}
int main() {
  ios::sync_with_stdio(0);
  long long n, k;
  cin >> n >> k;
  long long rs[k + 1];
  long long c[k + 1];
  long long prod = mpow(2, n);
  rs[0] = mpow(2, n);
  fill(c, c + k + 1, 0);
  c[0] = 1;
  for (long long i = 1; i <= k; ++i) {
    prod = (prod * (n - i + 1)) % mod;
    prod = (prod * 500000004) % mod;
    long long cnew[k + 1];
    cnew[0] = c[0];
    for (long long j = k; j > 0; --j) {
      cnew[j] = c[j - 1];
    }
    cnew[0] = 0;
    for (long long j = k; j >= 0; --j) {
      cnew[j] = (cnew[j] + c[j] - (i * c[j]) % mod) % mod;
      if (cnew[j] < 0) cnew[j] += mod;
    }
    for (long long j = 0; j <= k; ++j) c[j] = cnew[j];
    rs[i] = prod;
    for (long long j = i - 1; j >= 0; --j) {
      rs[i] = (rs[i] - (rs[j] * c[j]) % mod) % mod;
      if (rs[i] < 0) rs[i] += mod;
    }
  }
  cout << (rs[k] % mod) << endl;
  return 0;
}