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 maxn = 5005;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const double eps = 1e-9;
int n, k;
long long s[maxn][maxn], P[maxn], _pow[maxn];
long long fast_mod(long long a, long long n, long long Mod) {
  long long ans = 1;
  a %= Mod;
  while (n) {
    if (n & 1) ans = (ans * a) % Mod;
    a = (a * a) % Mod;
    n >>= 1;
  }
  return ans;
}
void init() {
  for (int i = 1; i < maxn; i++) {
    s[i][0] = 0, s[i][i] = 1;
    for (int j = 1; j < i; j++) {
      s[i][j] = (s[i - 1][j] * j % mod + s[i - 1][j - 1]) % mod;
    }
  }
  P[0] = 1;
  for (int i = 1; i <= k; i++) P[i] = P[i - 1] * (n - i + 1) % mod;
  long long inv = fast_mod(2, mod - 2, mod);
  _pow[0] = fast_mod(2, n, mod);
  for (int i = 1; i <= k; i++) _pow[i] = _pow[i - 1] * inv % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  init();
  long long ans = 0;
  for (int i = 1; i <= k; i++) {
    ans += (s[k][i] * P[i]) % mod * _pow[i] % mod;
    ans %= mod;
  }
  printf("%lld\n", ans);
}
