#include <bits/stdc++.h>
using namespace std;
using ll = long long;

#define rep(i,s,n) for(ll i = (s); i < (n); i++)
#define rep0(i,n) rep(i,0,n)

#define BR "\n"
#define SP " "
#define fcout cout << fixed << setprecision(18)

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

  int q; cin >> q;
  rep0(i, q) {
    double x, y;
    cin >> x >> y;

    double mx, my;
    if (x1 == x2) {
      mx = x1;
      my = y;
    } else if (y1 == y2) { 
      mx = x;
      my = y1;
    } else {
      double a1 = (y2 - y1) / (x2 - x1);
      double b1 = y1 - a1 * x1;
      double a2 = -1 / a1;
      double b2 = y - a2 * x;

      mx = -(b1 - b2) / (a1 - a2);
      my = mx * a1 + b1;
    }

    double ansx, ansy;
    ansx = mx - (x - mx);
    ansy = my - (y - my);
    fcout << ansx << SP << ansy << BR;
  }

  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 <bits/stdc++.h>
using namespace std;
using ll = long long;

#define rep(i,s,n) for(ll i = (s); i < (n); i++)
#define rep0(i,n) rep(i,0,n)

#define BR "\n"
#define SP " "
#define fcout cout << fixed << setprecision(18)

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

  int q; cin >> q;
  rep0(i, q) {
    double x, y;
    cin >> x >> y;

    double mx, my;
    if (x1 == x2) {
      mx = x1;
      my = y;
    } else if (y1 == y2) { 
      mx = x;
      my = y1;
    } else {
      double a1 = (y2 - y1) / (x2 - x1);
      double b1 = y1 - a1 * x1;
      double a2 = -1 / a1;
      double b2 = y - a2 * x;

      mx = -(b1 - b2) / (a1 - a2);
      my = mx * a1 + b1;
    }

    double ansx, ansy;
    ansx = mx - (x - mx);
    ansy = my - (y - my);
    fcout << ansx << SP << ansy << BR;
  }

  return 0;
}

```