#include <bits/stdc++.h>
using namespace std;
const int maxk = 5e3 + 10;
const int mod = 1e9 + 7;
int st[maxk][maxk];
int pw(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return ans;
}
int main() {
  st[0][0] = 1;
  for (int i = 1; i < maxk; i++)
    for (int j = 1; j < maxk; j++)
      st[i][j] = (st[i - 1][j - 1] + 1ll * st[i - 1][j] * j) % mod;
  int n, k;
  scanf("%d%d", &n, &k);
  int ans = 0;
  for (int i = 0; i <= min(k, n); i++) {
    int now = 1;
    for (int j = n - i + 1; j <= n; j++) now = 1ll * now * j % mod;
    ans = (ans + 1ll * st[k][i] * now % mod * pw(2, n - i)) % mod;
  }
  printf("%d", ans);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 maxk = 5e3 + 10;
const int mod = 1e9 + 7;
int st[maxk][maxk];
int pw(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return ans;
}
int main() {
  st[0][0] = 1;
  for (int i = 1; i < maxk; i++)
    for (int j = 1; j < maxk; j++)
      st[i][j] = (st[i - 1][j - 1] + 1ll * st[i - 1][j] * j) % mod;
  int n, k;
  scanf("%d%d", &n, &k);
  int ans = 0;
  for (int i = 0; i <= min(k, n); i++) {
    int now = 1;
    for (int j = n - i + 1; j <= n; j++) now = 1ll * now * j % mod;
    ans = (ans + 1ll * st[k][i] * now % mod * pw(2, n - i)) % mod;
  }
  printf("%d", ans);
  return 0;
}

```