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, MOD = 1000000007, INV2 = 500000004;
template <typename T>
void read(T &x) {
  bool neg = false;
  unsigned char c = getchar();
  for (; (c ^ 48) > 9; c = getchar())
    if (c == '-') neg = true;
  for (x = 0; (c ^ 48) < 10; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (neg) x = -x;
}
int n, k, S[N][N], ans;
int pow(int aa, int b) {
  long long c = 1, a = aa;
  while (b) {
    if (b & 1) (c *= a) %= MOD;
    b >>= 1;
    (a *= a) %= MOD;
  }
  return (int)c;
}
int main() {
  long long t = 1, s = 1;
  read(n);
  read(k);
  S[0][0] = 1;
  for (int i = 1, i_end = k + 1; i < i_end; ++i)
    for (int j = 1, j_end = k + 1; j < j_end; ++j) {
      S[i][j] = (S[i - 1][j - 1] + j * (long long)S[i - 1][j] % MOD) % MOD;
    }
  s = pow(2, n);
  for (int i = 0, i_end = k + 1; i < i_end; ++i) {
    (ans += S[k][i] * s % MOD * t % MOD) %= MOD;
    s = s * INV2 % MOD;
    (t *= n - i) %= MOD;
  }
  printf("%d\n", ans);
  return 0;
}
