#include<bits/stdc++.h>
typedef long long ll;
typedef long double ld;
using namespace std;
#define y1 ________y1
ld x1,x2,xp,y1,y2,yp,slope,ansx,ansy;
ll Q;
bool flx,fly;
signed main(){
	scanf("%Lf%Lf%Lf%Lf",&x1,&y1,&x2,&y2);
	if (y1==y2) flx=true;
	else if(x1==x2) fly=true;
	else slope=(y2-y1)/(x2-x1);
	x2-=x1,y2-=y1;
	scanf("%lld",&Q);while (Q--){
		scanf("%Lf%Lf",&xp,&yp);
		if(flx) printf("%.10Lf %.10Lf\n",xp*2-xp,y1*2-yp);
		else if(fly) printf("%.10Lf %.10Lf\n",x1*2-xp,yp*2-yp);
		else{
			ansx=(yp*x2*y2+xp*x2*x2-y1*y2*x2+x1*y2*y2)/(y2*y2+x2*x2);
			ansy=(-x2*ansx+yp*y2+xp*x2)/y2;
			printf("%.10Lf %.10Lf\n",ansx*2-xp,ansy*2-yp);
		}
	}
	fclose(stdin);fclose(stdout);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<bits/stdc++.h>
typedef long long ll;
typedef long double ld;
using namespace std;
#define y1 ________y1
ld x1,x2,xp,y1,y2,yp,slope,ansx,ansy;
ll Q;
bool flx,fly;
signed main(){
	scanf("%Lf%Lf%Lf%Lf",&x1,&y1,&x2,&y2);
	if (y1==y2) flx=true;
	else if(x1==x2) fly=true;
	else slope=(y2-y1)/(x2-x1);
	x2-=x1,y2-=y1;
	scanf("%lld",&Q);while (Q--){
		scanf("%Lf%Lf",&xp,&yp);
		if(flx) printf("%.10Lf %.10Lf\n",xp*2-xp,y1*2-yp);
		else if(fly) printf("%.10Lf %.10Lf\n",x1*2-xp,yp*2-yp);
		else{
			ansx=(yp*x2*y2+xp*x2*x2-y1*y2*x2+x1*y2*y2)/(y2*y2+x2*x2);
			ansy=(-x2*ansx+yp*y2+xp*x2)/y2;
			printf("%.10Lf %.10Lf\n",ansx*2-xp,ansy*2-yp);
		}
	}
	fclose(stdin);fclose(stdout);return 0;
}

```