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 long long M = 1e9 + 7;
const int MAXN = 5005;
long long n, k;
long long mem[MAXN][MAXN];
long long fact[MAXN];
long long modpow(long long b, long long p) {
  if (p <= 0) return 1;
  long long a = modpow(b, p / 2);
  return p % 2 ? a * a % M * b % M : a * a % M;
}
long long inv(long long b) { return modpow(b, M - 2); }
long long dp(int i, int need) {
  if (mem[i][need] != -1) return mem[i][need];
  if (i == 0) return !need;
  return mem[i][need] =
             (dp(i - 1, need) * need + dp(i - 1, need - 1) * need) % M;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  memset(mem, -1, sizeof mem);
  long long ans = 0;
  for (int j = 1; j <= k; j++) {
    long long dem = 1;
    long long num = 1;
    for (long long i = n, k = 0; k < j; k++, i--) dem = (dem * i) % M;
    for (long long i = 1; i <= j; i++) num = (num * i) % M;
    ans = (ans + dem * inv(num) % M * modpow(2, n - j) % M * dp(k, j) % M) % M;
  }
  cout << ans << endl;
}
