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 p = 1000000007;
long long pow_mod(long long x, long long n) {
  long long s = 1;
  for (; n; n >>= 1) {
    if (n & 1) s = s * x % p;
    x = x * x % p;
  }
  return s;
}
int n, K;
const int N = 5000;
long long S[5010][5010];
void init() {
  S[0][0] = 1;
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= N; ++j)
      S[i][j] = (S[i - 1][j - 1] + j * S[i - 1][j]) % p;
}
long long mul = 1, ans;
int main() {
  cin >> n >> K;
  init();
  for (int i = 0; i <= min(n, K); mul = mul * (n - i) % p, ++i)
    ans = (ans + S[K][i] * mul % p * pow_mod(2, n - i)) % p;
  cout << (ans + p) % p << endl;
  return 0;
}
