Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long d = 0, f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    d = d * 10 + s - '0';
    s = getchar();
  }
  return d * f;
}
long long ksm(long long x, long long y) {
  long long s = 1;
  while (y) {
    if (y & 1) (s *= x) %= 1000000007;
    (x *= x) %= 1000000007;
    y >>= 1;
  }
  return s;
}
long long s[5005][5005];
int main() {
  long long n = read(), k = read();
  s[0][0] = 1;
  for (long long i = 1; i <= k; i++)
    for (long long j = 1; j <= i; j++)
      s[i][j] = (s[i - 1][j - 1] + s[i - 1][j] * j % 1000000007) % 1000000007;
  long long p = ksm(2, n), pp = 1, ans = 0, inv = ksm(2, 1000000007 - 2);
  for (long long i = 0; i <= k; i++) {
    (ans += s[k][i] * pp % 1000000007 * p % 1000000007) %= 1000000007;
    (p *= inv) %= 1000000007;
    (pp *= (n - i + 1000000007) % 1000000007) %= 1000000007;
  }
  if (!k) ans = (ans - 1 + 1000000007) % 1000000007;
  cout << ans;
  return 0;
}