#include <iostream>
#include <iomanip>
using namespace std;
int main(void){
    struct Point {
        double x;
        double y;
    };
    Point p1, p2, p;
    double a, b, X, Y;
    int q;
    cin >> p1.x >> p1.y;
    cin >> p2.x >> p2.y;
    cin >> q;

    if(p1.x == p2.x){
        for(int i = 0; i < q; i++){
            cin >> p.x >> p.y;
            /*もしp1とp2のx座標が等しければ、点pと点Xは一致するため、
            点Xのx座標は線分p1p2の中点でy座標は点pのy座標である*/
            cout << fixed << setprecision(8) << 2 * p1.x - p.x << " ";
            cout << fixed << setprecision(8) << p.y << endl;
        }
    }
    else{
        double l = (p2.y - p1.y) / (p2.x - p1.x); //直線p1p2の傾き
        for(int i = 0; i < q; i++){
            cin >> p.x >> p.y;
            a = l * ((p.y - p1.y) - l * (p.x - p1.x)) / (1 + l * l) + p.x;
            b = (l * (l * p.y + p.x - p1.x) + p1.y) / (1 + l * l);
            X = 2 * a - p.x;
            Y = 2 * b - p.y;

            cout << fixed << setprecision(8) << X << " ";
            cout << fixed << setprecision(8) << Y << endl;
        }
    }
    return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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>
using namespace std;
int main(void){
    struct Point {
        double x;
        double y;
    };
    Point p1, p2, p;
    double a, b, X, Y;
    int q;
    cin >> p1.x >> p1.y;
    cin >> p2.x >> p2.y;
    cin >> q;

    if(p1.x == p2.x){
        for(int i = 0; i < q; i++){
            cin >> p.x >> p.y;
            /*もしp1とp2のx座標が等しければ、点pと点Xは一致するため、
            点Xのx座標は線分p1p2の中点でy座標は点pのy座標である*/
            cout << fixed << setprecision(8) << 2 * p1.x - p.x << " ";
            cout << fixed << setprecision(8) << p.y << endl;
        }
    }
    else{
        double l = (p2.y - p1.y) / (p2.x - p1.x); //直線p1p2の傾き
        for(int i = 0; i < q; i++){
            cin >> p.x >> p.y;
            a = l * ((p.y - p1.y) - l * (p.x - p1.x)) / (1 + l * l) + p.x;
            b = (l * (l * p.y + p.x - p1.x) + p1.y) / (1 + l * l);
            X = 2 * a - p.x;
            Y = 2 * b - p.y;

            cout << fixed << setprecision(8) << X << " ";
            cout << fixed << setprecision(8) << Y << endl;
        }
    }
    return 0;
}

```