Problem: For given three points p1, p2, p, find the reflection point x of p onto p1p2.

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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

Solution (in CPP):

Solution:
#include<iostream>
#include<cmath>
#include<cstdio>
#define EPS (1e-10)
#define equal(a,b)  fabs((a)-(b))<EPS
using namespace std;
struct Point{
	double x,y;
	Point(double _x,double _y):x(_x),y(_y){  }
	Point operator + (const Point &a){
		return Point(x+a.x,y+a.y);
	}
	Point operator - (const Point &a){
		return Point(x-a.x,y-a.y);
	}
	Point operator *(double k){
		return Point (k*x,k*y);
	}
	Point operator /(double k){
		return Point(x/k,y/k);
	}
	
	friend double norm(const Point &a){
		return a.x*a.x+a.y*a.y;
	}
	friend double abs(const Point &a){
		return sqrt(norm(a));
	}
	
	friend double dot(const Point &a,const Point &b){
		return a.x*b.x+a.y*b.y;
	}
	friend double cross(const Point &a,const Point &b){
		return fabs(a.x*b.y-b.x*a.y);
	}
};
typedef Point Vector;
bool isOrthogonal(const Vector &a,const Vector & b){
	return equal(cross(a,b),0.0);
}
bool isParallel(const Vector &a,const Vector &b){
	return equal(dot(a,b),0.0);
}

int main(){
	int x1,x2,y1,y2,q;
	cin>>x1>>y1>>x2>>y2>>q;
	Point a(x1,y1),b(x2,y2);
	while(q--){
		int x3,y3;
		cin>>x3>>y3;
		Point c(x3,y3);
		double t=dot(c-a,b-a)/abs(b-a);
		Vector det= (b-a) / abs(b-a) *t;
		Point ort=a+det;
		det=ort-c;
		ort=c+det*2;
		printf("%.10f %.10f\n",ort.x,ort.y);
	}
	return 0;
}