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 mod = 1e9 + 7;
map<pair<int, int>, int> mp;
int g[5005][5005];
int n, m;
int expmod(int a, int b, int mod) {
  int res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = 1ll * res * a % mod;
    a = 1ll * a * a % mod;
  }
  return res;
}
int func(int x, int y) {
  if (y == 0) return expmod(2, x, mod);
  if (x == 0) return 0;
  pair<int, int> rpt = pair<int, int>(x, y);
  int ans = 1;
  ans = 1ll * func(x - 1, y - 1) * x % mod;
  return ans;
}
void init() {
  g[0][0] = 1;
  for (int i = 1; i < 5005; i++) {
    for (int k = 1; k <= i; k++)
      g[i][k] = (1ll * k * g[i - 1][k] + g[i - 1][k - 1]) % mod;
  }
}
int main() {
  init();
  scanf("%d %d", &m, &n);
  int ans = 0;
  for (int k = 0; k <= n; k++) {
    ans = (ans + 1ll * g[n][k] * func(m, k)) % mod;
  }
  printf("%d\n", ans);
}
