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 = 2e5 + 10;
const int mod = 1e9 + 7;
const int INF = 1e9 + 10;
const double EPS = 1e-8;
long long n, m;
long long x[10005], y[10005];
long long power(long long a, long long b) {
  long long c = 1;
  while (b) {
    if (b % 2 == 1) c = c * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return c;
}
int main() {
  scanf("%lld%lld", &n, &m);
  memset(x, 0, sizeof(x));
  long long A = n, ans = 0;
  x[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j <= n - A; j++) {
      if (A > 0)
        y[j] = (A * x[j] % mod + x[j] * j % mod + x[j + 1] * (j + 1)) % mod;
      else
        y[j] = x[j + 1] * (j + 1) % mod;
    }
    if (i == m) {
      long long sum = 0;
      for (int j = 0; j <= n - A; j++) {
        sum += y[j];
        if (sum >= mod) sum -= mod;
        if (sum <= -mod) sum += mod;
      }
      ans = power(2, A - 1) * sum % mod;
    }
    for (int j = n - A; j >= 0; j--) x[j + 1] = y[j];
    x[0] = 0;
    A--;
  }
  printf("%lld\n", ans);
  return 0;
}
