#include <iostream>
#include <math.h>
#include <cstdio>
using namespace std;

int main(int argc, const char * argv[]) {
    double xp1, yp1, xp2, yp2;
    int q;
    double xp, yp;
    
    cin >> xp1 >> yp1 >> xp2 >> yp2;
    cin >> q;
    
    double dist = sqrt((xp1 - xp2) * (xp1 - xp2) + (yp1 - yp2) * (yp1 -yp2));
    double nx, ny; //unit vector along the line
    nx = (xp2 - xp1) / dist;
    ny = (yp2 - yp1) / dist;
    
    for (int i = 0; i < q; ++i) {
        cin >> xp >> yp;
        double t = nx * (xp - xp1) + ny * (yp - yp1);
        printf("%.10lf %.10lf\n", 2 * (xp1 + nx * t) - xp, 2 * (yp1 + ny * t) - yp);
    }
    
    return 0;
}


### Prompt

Generate 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 <math.h>
#include <cstdio>
using namespace std;

int main(int argc, const char * argv[]) {
    double xp1, yp1, xp2, yp2;
    int q;
    double xp, yp;
    
    cin >> xp1 >> yp1 >> xp2 >> yp2;
    cin >> q;
    
    double dist = sqrt((xp1 - xp2) * (xp1 - xp2) + (yp1 - yp2) * (yp1 -yp2));
    double nx, ny; //unit vector along the line
    nx = (xp2 - xp1) / dist;
    ny = (yp2 - yp1) / dist;
    
    for (int i = 0; i < q; ++i) {
        cin >> xp >> yp;
        double t = nx * (xp - xp1) + ny * (yp - yp1);
        printf("%.10lf %.10lf\n", 2 * (xp1 + nx * t) - xp, 2 * (yp1 + ny * t) - yp);
    }
    
    return 0;
}


```