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>
const int64_t MOD = 1e9 + 7;
using namespace std;
int64_t n, k, dp[5005][5005];
int64_t power(int64_t x, int64_t y) {
  int64_t res = 1;
  x = x % MOD;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return res % MOD;
}
int64_t solve(int64_t k, int64_t a, int64_t b) {
  if (dp[k][a] != -1) return dp[k][a];
  if (k == 0) return dp[k][a] = power(2, b);
  return dp[k][a] =
             (a * solve(k - 1, a, b) + b * solve(k - 1, a + 1, b - 1)) % MOD;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int64_t n, k;
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  cout << solve(k, 0, n) << endl;
  return 0;
}
