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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long MOD = 1000000007;
const int INF = 1000000007;
const long long LINF = 1LL * INF * INF;
const int MAXN = 5007;
const double EPS = 1e-6;
const int MAXK = 5007, R2 = 500000004;
long long A[MAXK][MAXK];
int main() {
  int n, k, n1;
  long long ans = 0, c = 1, a = 2;
  cin >> n >> k;
  A[1][1] = 1;
  for (int i = (1); i < (k); ++i) {
    for (int j = (1); j < (i + 1); ++j) {
      A[i + 1][j + 1] += A[i][j];
      A[i + 1][j + 1] %= MOD;
      A[i + 1][j] += j * A[i][j];
      A[i + 1][j] %= MOD;
    }
  }
  n1 = n;
  while (n1) {
    if (n1 & 1) {
      c *= a;
      c %= MOD;
    }
    a = (a * a) % MOD;
    n1 >>= 1;
  }
  for (int i = (1); i < (k + 1); ++i) {
    c *= (n - i + 1LL);
    c %= MOD;
    c *= R2;
    c %= MOD;
    ans += A[k][i] * c;
    ans %= MOD;
  }
  cout << ans;
  return 0;
}
