#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long fexp(long long x, long long t) {
  long long ans = 1, base = x;
  while (t) {
    if (t % 2) ans = ans * base % mod;
    base = base * base % mod;
    t /= 2;
  }
  return ans;
}
long long dp[5001][5001];
int cnt = 0;
long long dfs(long long a, long long b, long long c) {
  if (dp[a][b] != -1) return dp[a][b];
  if (a == 0) return dp[a][b] = fexp(2, c);
  return dp[a][b] = (b * dfs(a - 1, b, c) + c * dfs(a - 1, b + 1, c - 1)) % mod;
}
int main() {
  memset(dp, -1, sizeof dp);
  long long n, k;
  scanf("%lld%lld", &n, &k);
  printf("%lld\n", dfs(k, 0, n));
  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 long long mod = 1e9 + 7;
long long fexp(long long x, long long t) {
  long long ans = 1, base = x;
  while (t) {
    if (t % 2) ans = ans * base % mod;
    base = base * base % mod;
    t /= 2;
  }
  return ans;
}
long long dp[5001][5001];
int cnt = 0;
long long dfs(long long a, long long b, long long c) {
  if (dp[a][b] != -1) return dp[a][b];
  if (a == 0) return dp[a][b] = fexp(2, c);
  return dp[a][b] = (b * dfs(a - 1, b, c) + c * dfs(a - 1, b + 1, c - 1)) % mod;
}
int main() {
  memset(dp, -1, sizeof dp);
  long long n, k;
  scanf("%lld%lld", &n, &k);
  printf("%lld\n", dfs(k, 0, n));
  return 0;
}

```