#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int Maxk = 5010;
long long n, k;
long long Pow(long long x, long long y) {
  if (!y) return 1;
  if (y == 1) return x;
  long long t = Pow(x, y >> 1), ans = t * t % mod;
  if (y & 1) ans = ans * x % mod;
  return ans;
}
int f[Maxk][Maxk];
int main() {
  scanf("%d%d", &n, &k);
  f[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++)
      f[i][j] = ((long long)f[i - 1][j] * j +
                 (long long)f[i - 1][j - 1] * (n - j + 1)) %
                mod;
    f[0][0] = 0;
  }
  long long ans = 0;
  for (int i = 1; i <= min(n, k); i++)
    ans = (ans + f[k][i] * Pow(2LL, n - i) % mod) % mod;
  printf("%I64d", ans);
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int Maxk = 5010;
long long n, k;
long long Pow(long long x, long long y) {
  if (!y) return 1;
  if (y == 1) return x;
  long long t = Pow(x, y >> 1), ans = t * t % mod;
  if (y & 1) ans = ans * x % mod;
  return ans;
}
int f[Maxk][Maxk];
int main() {
  scanf("%d%d", &n, &k);
  f[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++)
      f[i][j] = ((long long)f[i - 1][j] * j +
                 (long long)f[i - 1][j - 1] * (n - j + 1)) %
                mod;
    f[0][0] = 0;
  }
  long long ans = 0;
  for (int i = 1; i <= min(n, k); i++)
    ans = (ans + f[k][i] * Pow(2LL, n - i) % mod) % mod;
  printf("%I64d", ans);
}

```