#include <iostream>
#include <iomanip>
#include <cmath>

typedef long double ld;

ld dissq(double x1, double y1, double x2, double y2)
{
	return (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2);
}

int main()
{
	ld x1, y1, x2, y2;
	std::cin >> x1 >> y1 >> x2 >> y2;
	int q; std::cin >> q;
	for (int i=0; i<q; i++) {
		ld x0, y0;
		std::cin >> x0 >> y0;
		ld s = (dissq(x0,y0,x1,y1) + dissq(x1,y1,x2,y2) - dissq(x0,y0,x2,y2)) / (2.0 * dissq(x1,y1,x2,y2));
		ld x = (1.0 - s) * x1 + s * x2;
		ld y = (1.0 - s) * y1 + s * y2;
		std::cout << std::setprecision(8) << std::fixed << 2.0 * x - x0 << " " << 2.0 * y - y0 << std::endl;
	}
	return 0;
}


### Prompt

Please formulate 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>
#include <cmath>

typedef long double ld;

ld dissq(double x1, double y1, double x2, double y2)
{
	return (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2);
}

int main()
{
	ld x1, y1, x2, y2;
	std::cin >> x1 >> y1 >> x2 >> y2;
	int q; std::cin >> q;
	for (int i=0; i<q; i++) {
		ld x0, y0;
		std::cin >> x0 >> y0;
		ld s = (dissq(x0,y0,x1,y1) + dissq(x1,y1,x2,y2) - dissq(x0,y0,x2,y2)) / (2.0 * dissq(x1,y1,x2,y2));
		ld x = (1.0 - s) * x1 + s * x2;
		ld y = (1.0 - s) * y1 + s * y2;
		std::cout << std::setprecision(8) << std::fixed << 2.0 * x - x0 << " " << 2.0 * y - y0 << std::endl;
	}
	return 0;
}


```