#include<cstdio>
#include<cmath>

using namespace std;

int main()
{
	int x0, y0, x1, y1, q, x2, y2;
	scanf("%d %d %d %d", &x0, &y0, &x1, &y1);
	scanf("%d", &q);
	for (int i = 0; i < q; i++)
	{
		scanf("%d %d", &x2, &y2);
		double x, y;
		x = (double)x0 + 1.0*(x1-x0)*((x1 - x0)*(x2 - x0) + (y1 - y0)*(y2 - y0)) / (pow(x1 - x0, 2) + pow(y1 - y0, 2));
		y = (double)y0 + 1.0*(y1-y0)*((x1 - x0)*(x2 - x0) + (y1 - y0)*(y2 - y0)) / (pow(x1 - x0, 2) + pow(y1 - y0, 2));
		double xr, yr;
		xr = 2 * x - x2;
		yr = 2 * y - y2;
		printf("%.10lf %.10lf\n", xr, yr);
	}
	return 0;
}
### Prompt

Develop a solution in CPP 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<cstdio>
#include<cmath>

using namespace std;

int main()
{
	int x0, y0, x1, y1, q, x2, y2;
	scanf("%d %d %d %d", &x0, &y0, &x1, &y1);
	scanf("%d", &q);
	for (int i = 0; i < q; i++)
	{
		scanf("%d %d", &x2, &y2);
		double x, y;
		x = (double)x0 + 1.0*(x1-x0)*((x1 - x0)*(x2 - x0) + (y1 - y0)*(y2 - y0)) / (pow(x1 - x0, 2) + pow(y1 - y0, 2));
		y = (double)y0 + 1.0*(y1-y0)*((x1 - x0)*(x2 - x0) + (y1 - y0)*(y2 - y0)) / (pow(x1 - x0, 2) + pow(y1 - y0, 2));
		double xr, yr;
		xr = 2 * x - x2;
		yr = 2 * y - y2;
		printf("%.10lf %.10lf\n", xr, yr);
	}
	return 0;
}
```