#include <iostream>
#include <utility>
#include <cmath>
#include <stdio.h>

using namespace std;

int main(void){

    pair<double, double> p0, p1, p, ans;
    int n;
    cin >> p0.first >> p0.second >> p1.first >> p1.second;
    cin >> n;

    while(n--){
        cin >> p.first >> p.second;
        double a;
        if(p1.second - p0.second == 0){
            ans.first = p.first;
            ans.second = p1.second;
        }else if(p1.first - p0.first == 0){
            ans.first = p1.first;
            ans.second = p.second;
        }else{
            a = (p1.second - p0.second) / (p1.first - p0.first);
            ans.first = (a / (a*a + 1) ) * (-p0.second + p.second + a*p0.first + p.first/a);
            ans.second = (-1/a) * (ans.first - p.first) + p.second;
        }
        printf("%.10f ", ans.first * 2 - p.first);
        printf("%.10f", ans.second * 2 - p.second);
        cout << 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 <utility>
#include <cmath>
#include <stdio.h>

using namespace std;

int main(void){

    pair<double, double> p0, p1, p, ans;
    int n;
    cin >> p0.first >> p0.second >> p1.first >> p1.second;
    cin >> n;

    while(n--){
        cin >> p.first >> p.second;
        double a;
        if(p1.second - p0.second == 0){
            ans.first = p.first;
            ans.second = p1.second;
        }else if(p1.first - p0.first == 0){
            ans.first = p1.first;
            ans.second = p.second;
        }else{
            a = (p1.second - p0.second) / (p1.first - p0.first);
            ans.first = (a / (a*a + 1) ) * (-p0.second + p.second + a*p0.first + p.first/a);
            ans.second = (-1/a) * (ans.first - p.first) + p.second;
        }
        printf("%.10f ", ans.first * 2 - p.first);
        printf("%.10f", ans.second * 2 - p.second);
        cout << endl;
    }

    return 0;
}
```