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:
//
// Created by 肖龙 on 2020/2/20.
//


/*
 * 求点和直线的投影
 */
#include<iostream>
#include <cmath>
#include <iomanip>

using namespace std;

typedef struct point{
    double x,y;
}Point,Vector;

typedef struct line{
    Point x;
    Vector k;
}Line;

double dot(Vector x,Vector y){
    return x.x*y.x+x.y*y.y;
}

Vector getV(Point x,Point y){
    Vector v;
    v.x=y.x-x.x;
    v.y=y.y-x.y;
    return v;
}
//点在直线的投影
Point Projection(Line x,Point y){
    Point answ;
    answ = x.x;

    double len = dot(x.k,getV(x.x,y));
    len/=sqrt(x.k.x*x.k.x+x.k.y*x.k.y);
    Vector temp  = x.k;
    temp.x /=sqrt(x.k.x*x.k.x+x.k.y*x.k.y);
    temp.y/=sqrt(x.k.x*x.k.x+x.k.y*x.k.y);
    answ.x+=temp.x*len;
    answ.y+=temp.y*len;
    return answ;
}
//点与直线的对称点
Point Reflection(Line x,Point y){
    Point half = Projection(x,y);
    Point answ;
    answ.x=y.x+2*(half.x-y.x);
    answ.y=y.y+2*(half.y-y.y);
    return answ;
}

int main(){
    Point p1,p2,p;
    cin>>p1.x>>p1.y>>p2.x>>p2.y;
    int q;
    cin>>q;
    Vector v2;
    v2 = getV(p1,p2);
    Line l;
    l.k=v2;
    l.x=p1;
    for(int i=1;i<=q;i++){
        cin>>p.x>>p.y;
        Point answ = Reflection(l,p);
        cout<<setprecision(10)<<setiosflags(ios::fixed)<<answ.x<<" "<<answ.y<<endl;
    }
    return 0;
}