#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;

int main(){
	double x1,x2,xp,y1,y2,yp,slope,ansx,ansy;
	int q;
	bool X_FLG = false,Y_FLG = false;

	scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
	scanf("%d",&q);

	if(y1 == y2){
		X_FLG = true;
	}else if(x1 == x2){
		Y_FLG = true;
	}else{
		slope = (y2-y1)/(x2-x1);
	}

	for(int i=0;i<q;i++){
		scanf("%lf %lf",&xp,&yp);
		if(X_FLG){
			ansx = xp,ansy=y1;
			printf("%.10lf %.10lf\n",2*ansx-xp,2*ansy-yp);
		}else if(Y_FLG){
			ansx = x1,ansy = yp;
			printf("%.10lf %.10lf\n",2*ansx-xp,2*ansy-yp);
		}else{
			ansx = (yp*(x2-x1)*(y2-y1)+xp*(x2-x1)*(x2-x1)-y1*(y2-y1)*(x2-x1)+x1*(y2-y1)*(y2-y1))/((y2-y1)*(y2-y1)+(x2-x1)*(x2-x1));
			ansy = ((x1-x2)*ansx+yp*(y2-y1)+xp*(x2-x1))/(y2-y1);
			printf("%.10lf %.10lf\n",2*ansx-xp,2*ansy-yp);
		}

	}

	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 <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;

int main(){
	double x1,x2,xp,y1,y2,yp,slope,ansx,ansy;
	int q;
	bool X_FLG = false,Y_FLG = false;

	scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
	scanf("%d",&q);

	if(y1 == y2){
		X_FLG = true;
	}else if(x1 == x2){
		Y_FLG = true;
	}else{
		slope = (y2-y1)/(x2-x1);
	}

	for(int i=0;i<q;i++){
		scanf("%lf %lf",&xp,&yp);
		if(X_FLG){
			ansx = xp,ansy=y1;
			printf("%.10lf %.10lf\n",2*ansx-xp,2*ansy-yp);
		}else if(Y_FLG){
			ansx = x1,ansy = yp;
			printf("%.10lf %.10lf\n",2*ansx-xp,2*ansy-yp);
		}else{
			ansx = (yp*(x2-x1)*(y2-y1)+xp*(x2-x1)*(x2-x1)-y1*(y2-y1)*(x2-x1)+x1*(y2-y1)*(y2-y1))/((y2-y1)*(y2-y1)+(x2-x1)*(x2-x1));
			ansy = ((x1-x2)*ansx+yp*(y2-y1)+xp*(x2-x1))/(y2-y1);
			printf("%.10lf %.10lf\n",2*ansx-xp,2*ansy-yp);
		}

	}

	return 0;
}

```