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<bits/stdc++.h>

using namespace std;
#define eps = (1e-10);
//点
struct Point{
    double x,y;
    Point(double _x=0,double _y=0):x(_x),y(_y){}
    Point operator + (Point p){return Point(x+p.x,y+p.y);}
    Point operator - (Point p){return Point(x-p.x,y-p.y);}
    Point operator * (double a){return Point(a*x,a*y);}
    Point operator / (double a){return Point(x/a,y/a);}
    double norm(){return x*x+y*y;}//模的平方
    double ABS() {return sqrt(norm());}//模
};
//线段
struct Segment{
    Point p1,p2;
};
//点积
double dot(Point a,Point b){
    return a.x*b.x+a.y*b.y;
}
//叉积
double cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
//p关于s的对称点
Point Reflection(Point p,Segment s){
    Point p1=s.p1;Point p2=s.p2;
    double A=(p1.y-p2.y);
    double B=-(p1.x-p2.x);
    double C=(p1.x-p2.x)*p1.y-p1.x*A;
    return {((B*B-A*A)*p.x-2*A*B*p.y-2*A*C)/(A*A+B*B),((A*A-B*B)*p.y-2*A*B*p.x-2*B*C)/(A*A+B*B)};
}
int main(){
    double x1,y1,x2,y2;
    scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
    Point p1={x1,y1};Point p2={x2,y2};
    Segment s={p1,p2};
    int T;scanf("%d",&T);
    double x,y;
    while(T--){
        scanf("%lf %lf",&x,&y);
        Point p={x,y};
        Point ans=Reflection(p,s);
        printf("%.11lf %.11lf\n",ans.x,ans.y);
    }
    return 0;
}