#include <iostream>
#include <iomanip>
#include <limits>

int main() {
  double x1, y1, x2, y2;
  std::cin >> x1 >> y1 >> x2 >> y2;

  int q;
  std::cin >> q;

  std::cout << std::fixed;
  std::cout << std::setprecision(std::numeric_limits<double>::max_digits10);
  for (int i = 0; i < q; ++i) {
      double x, y;
      std::cin >> x >> y;

      double num = -(x2 - x1) * (x1 - x) - (y2 - y1) * (y1 - y);
      double den = (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
      double t = num / den;
      
      double u = x1 + t * (x2 - x1);
      double v = y1 + t * (y2 - y1);

      double k = x + 2 * (u - x);
      double l = y + 2 * (v - y);

      std::cout << k << " " << l << std::endl;
  }

  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

For given three points p1, p2, p, find the reflection point x of p onto p1p2.

<image>

Constraints

* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* p1 and p2 are not identical.

Input


xp1 yp1 xp2 yp2
q
xp0 yp0
xp1 yp1
...
xpq−1 ypq−1


In the first line, integer coordinates of p1 and p2 are given. Then, q queries are given for integer coordinates of p.

Output

For each query, print the coordinate of the reflection point x. The output values should be in a decimal fraction with an error less than 0.00000001.

Examples

Input

0 0 2 0
3
-1 1
0 1
1 1


Output

-1.0000000000 -1.0000000000
0.0000000000 -1.0000000000
1.0000000000 -1.0000000000


Input

0 0 3 4
3
2 5
1 4
0 3


Output

4.2400000000 3.3200000000
3.5600000000 2.0800000000
2.8800000000 0.8400000000

### Response

```cpp
#include <iostream>
#include <iomanip>
#include <limits>

int main() {
  double x1, y1, x2, y2;
  std::cin >> x1 >> y1 >> x2 >> y2;

  int q;
  std::cin >> q;

  std::cout << std::fixed;
  std::cout << std::setprecision(std::numeric_limits<double>::max_digits10);
  for (int i = 0; i < q; ++i) {
      double x, y;
      std::cin >> x >> y;

      double num = -(x2 - x1) * (x1 - x) - (y2 - y1) * (y1 - y);
      double den = (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
      double t = num / den;
      
      double u = x1 + t * (x2 - x1);
      double v = y1 + t * (y2 - y1);

      double k = x + 2 * (u - x);
      double l = y + 2 * (v - y);

      std::cout << k << " " << l << std::endl;
  }

  return 0;
}

```