#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long n, k, ans, mult = 1, f[2][5010];
long long mul(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> k;
  f[1][1] = 1;
  for (int i = 2, v = 0; i <= k; i++, v ^= 1)
    for (int j = 1; j <= i; j++)
      f[v][j] = (f[v ^ 1][j] * j + f[v ^ 1][j - 1]) % mod;
  for (int i = 1; i <= min(k, n); i++) {
    mult = (mult * (n - i + 1)) % mod;
    ans = (ans + f[k & 1][i] * mult % mod * mul(2, n - i)) % mod;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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 long long mod = 1e9 + 7;
long long n, k, ans, mult = 1, f[2][5010];
long long mul(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> k;
  f[1][1] = 1;
  for (int i = 2, v = 0; i <= k; i++, v ^= 1)
    for (int j = 1; j <= i; j++)
      f[v][j] = (f[v ^ 1][j] * j + f[v ^ 1][j - 1]) % mod;
  for (int i = 1; i <= min(k, n); i++) {
    mult = (mult * (n - i + 1)) % mod;
    ans = (ans + f[k & 1][i] * mult % mod * mul(2, n - i)) % mod;
  }
  cout << ans << '\n';
  return 0;
}

```