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 N = 5e3 + 5;
const int mod = 1e9 + 7;
int S[N][N];
int n, k;
void init(int n) {
  for (int i = 1; i <= n; i++) S[i][1] = S[i][i] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 1; j <= i; j++)
      S[i][j] = (S[i - 1][j - 1] + 1ll * S[i - 1][j] * j % mod) % mod;
}
long long fac(int l, int r) {
  long long ans = 1;
  for (int i = l; i <= r; i++) ans = ans * i % mod;
  return ans;
}
long long pow(long long a, long long b, long long p) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) ans = ans * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return ans;
}
int main() {
  init(N - 4);
  scanf("%d%d", &n, &k);
  long long ans = 0;
  for (int i = 1; i <= min(k, n); i++) {
    long long now =
        1ll * S[k][i] * fac(n - i + 1, n) % mod * pow(2, n - i, mod) % mod;
    ans = (ans + now) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
