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;
int s2[5001][5001];
long long n, k, c[5001];
inline long long q(long long x, long long y) {
  long long sum = x, tmp = 1ll;
  while (y) {
    if (y & 1) tmp = tmp * sum % 1000000007;
    sum = sum * sum % 1000000007;
    y >>= 1;
  }
  return tmp;
}
int main() {
  cin >> n >> k;
  s2[0][0] = s2[1][1] = 1;
  for (int i = 1; i <= k; i++) s2[i][0] = 0;
  for (register int i = 1; i <= k; i++) {
    for (register int j = 1; j <= i; j++) {
      s2[i][j] =
          ((long long)s2[i - 1][j - 1] + 1ll * j * (long long)s2[i - 1][j]) %
          1000000007;
    }
  }
  c[0] = 1;
  for (long long i = 1; i <= k; i++)
    c[i] =
        c[i - 1] * (n - i + 1) % 1000000007 * q(i, 1000000007 - 2) % 1000000007;
  long long ans = 0, jie = 1, now = q(2, n), ni = q(2, 1000000007 - 2);
  for (register int i = 1; i <= min(n, k); i++) {
    jie = jie * (long long)i % 1000000007, now = now * ni % 1000000007;
    long long now2 = jie * (long long)s2[k][i] % 1000000007 * c[i] %
                     1000000007 * now % 1000000007;
    ans = (ans + now2) % 1000000007;
  }
  cout << ans;
}
