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 MAXN = 5e3 + 3, mod = 1e9 + 7;
int f[MAXN];
int n, k;
int pw(int, int);
void solve();
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  solve();
  return 0;
}
int pw(int a, int b) {
  if (b == 0) return 1;
  if (b & 1) return 1ll * pw(a, b - 1) * a % mod;
  int x = pw(a, b >> 1);
  return 1ll * x * x % mod;
}
void solve() {
  f[1] = n;
  for (int i = 2; i <= k; i++)
    for (int j = min(i, n); j > 0; j--)
      f[j] = ((1ll * f[j] * j) + (1ll * f[j - 1] * (n - j + 1))) % mod;
  long long ans = 0;
  for (int i = 1; i <= min(k, n); i++)
    (ans += 1ll * f[i] * pw(2, n - i)) %= mod;
  cout << ans << endl;
}
