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.
#include <bits/stdc++.h>
using namespace std;
const int K = 5000 + 10;
const int MOD = 1000000007;
inline void U(int& x, int y) {
  if ((x += y) >= MOD) x -= MOD;
}
inline int quickpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = (long long)x * x % MOD)
    if (y & 1) res = (long long)res * x % MOD;
  return res;
}
int n, k;
int dp[K][K];
int main() {
  scanf("%d%d", &n, &k);
  dp[0][0] = 1;
  int *f, *g;
  for (int i = 1; i <= k; ++i) {
    f = dp[i - 1], g = dp[i];
    for (int j = 0; j < i && j < n; ++j) {
      U(g[j + 1], (long long)f[j] * (n - j) % MOD);
      U(g[j], (long long)f[j] * j % MOD);
    }
    for (int j = n; j < i; ++j) U(g[j], (long long)f[j] * j % MOD);
  }
  int ans = 0;
  for (int i = 0; i <= k; ++i)
    U(ans, (long long)g[i] * quickpow(2, max(0, n - i)) % MOD);
  printf("%d", ans);
  return 0;
}
