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 INF = 0x3f3f3f3f, N = 5010, P = 1e9 + 7;
template <class T>
inline bool chkmin(T &A, T B) {
  return B < A ? A = B, 1 : 0;
}
template <class T>
inline bool chkmax(T &A, T B) {
  return A < B ? A = B, 1 : 0;
}
template <class T, class _>
inline void Add(T &first, _ second) {
  first += second;
  if (first >= P) first -= P;
  if (first < 0) first += P;
}
int n, k;
long long ans, S[N][N];
long long Pow(long long first, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * first % P;
    first = first * first % P;
    k >>= 1;
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &k);
  S[1][1] = n;
  for (int i = (2), i_end = (k); i <= i_end; ++i)
    for (int j = (1), j_end = (min(n, i)); j <= j_end; ++j)
      S[i][j] = (S[i - 1][j] * j + S[i - 1][j - 1] * (n - j + 1)) % P;
  for (int i = (1), i_end = (min(n, k)); i <= i_end; ++i)
    Add(ans, S[k][i] * Pow(2, n - i) % P);
  printf("%lld", ans);
  return 0;
}
