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 N = 5005;
const int MOD = 1e9 + 7;
int n, k;
long long stirling[N][N];
long long fall[N];
long long bigMod(long long a, int e) {
  if (e < 0) return 0;
  long long ret = 1;
  while (e) {
    if (e & 1) ret *= a, ret %= MOD;
    a *= a, a %= MOD, e >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> k;
  fall[0] = 1;
  stirling[0][0] = 1;
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= k; ++j) {
      stirling[i][j] =
          (j * 1LL * stirling[i - 1][j] + stirling[i - 1][j - 1]) % MOD;
    }
    if (i <= n + 1) fall[i] = fall[i - 1] * (n - i + 1LL) % MOD;
  }
  long long ans = 0;
  for (int i = 0; i <= k; ++i) {
    long long add = stirling[k][i] * fall[i] % MOD;
    add *= bigMod(2, n - i), add %= MOD;
    ans += add;
    if (ans >= MOD) ans -= MOD;
  }
  cout << ans << '\n';
  return 0;
}
