#include <bits/stdc++.h>
using namespace std;
int ali;
const long long MAXN = 5e3 + 10;
const long long Mod = 1e9 + 7;
long long dp[MAXN][MAXN];
long long pw(long long x, long long n) {
  ali++;
  if (n < 0) {
    ali++;
    return 0;
  }
  if (n == 0) {
    ali++;
    return 1;
  }
  ali++;
  long long w = pw(x, n / 2);
  w *= w;
  w %= Mod;
  ali++;
  if (n & 1) {
    ali++;
    w *= x;
    w %= Mod;
    ali++;
  }
  return w;
}
int main() {
  ali++;
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  ali++;
  long long n, k, res = 0;
  cin >> n >> k;
  ali++;
  dp[0][0] = 1;
  ali++;
  for (int i = 1; i <= k; i++) {
    ali++;
    for (int j = 1; j <= k; j++) {
      ali++;
      dp[i][j] = dp[i - 1][j] * j + dp[i - 1][j - 1] * (n - j + 1);
      dp[i][j] %= Mod;
      ali++;
    }
  }
  ali++;
  for (int i = 1; i <= k; i++) {
    ali++;
    res += dp[k][i] * pw(2, n - i);
    res %= Mod;
    ali++;
  }
  cout << res;
}

### Prompt

Please formulate 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;
int ali;
const long long MAXN = 5e3 + 10;
const long long Mod = 1e9 + 7;
long long dp[MAXN][MAXN];
long long pw(long long x, long long n) {
  ali++;
  if (n < 0) {
    ali++;
    return 0;
  }
  if (n == 0) {
    ali++;
    return 1;
  }
  ali++;
  long long w = pw(x, n / 2);
  w *= w;
  w %= Mod;
  ali++;
  if (n & 1) {
    ali++;
    w *= x;
    w %= Mod;
    ali++;
  }
  return w;
}
int main() {
  ali++;
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  ali++;
  long long n, k, res = 0;
  cin >> n >> k;
  ali++;
  dp[0][0] = 1;
  ali++;
  for (int i = 1; i <= k; i++) {
    ali++;
    for (int j = 1; j <= k; j++) {
      ali++;
      dp[i][j] = dp[i - 1][j] * j + dp[i - 1][j - 1] * (n - j + 1);
      dp[i][j] %= Mod;
      ali++;
    }
  }
  ali++;
  for (int i = 1; i <= k; i++) {
    ali++;
    res += dp[k][i] * pw(2, n - i);
    res %= Mod;
    ali++;
  }
  cout << res;
}

```