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>
template <class T>
T read(T &a) {
  a = 0;
  char x = getchar();
  bool f = 0;
  for (; x < '0' || x > '9'; x = getchar()) f |= x == '-';
  for (; x >= '0' && x <= '9'; x = getchar()) a = (a << 3) + (a << 1) + x - '0';
  if (f) a = -a;
  return a;
}
using namespace std;
const int N = 5005, MOD = 1e9 + 7;
int n, k;
long long f[N][N];
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return ans;
}
int main() {
  read(n), read(k);
  f[0][0] = 1;
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j <= k; ++j) {
      if (j) {
        f[i + 1][j] += f[i][j] * j % MOD;
        f[i + 1][j] %= MOD;
      }
      if (j < n) {
        f[i + 1][j + 1] += f[i][j] * (n - j) % MOD;
        f[i + 1][j + 1] %= MOD;
      }
    }
  }
  int lim = min(n, k);
  long long ans = 0;
  for (int i = 0; i <= lim; ++i) {
    ans = (ans + qpow(2, n - i) * f[k][i] % MOD) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
