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;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
#define rep(i,n) for(UL i=0; i<(n); i++)

struct Vec{
 double x,y;
 Vec(double xx, double yy) : x(xx),y(yy) {}
 Vec operator+() const {return Vec(x,y);}
 Vec operator-() const {return Vec(-x,-y);}
 Vec operator+(const Vec& r) const {return Vec(x+r.x,y+r.y);}
 Vec operator-(const Vec& r) const {return Vec(x-r.x,y-r.y);}
 Vec operator*(double r) const {return Vec(x*r,y*r);}
 Vec operator/(double r) const {return Vec(x/r,y/r);}
 Vec& operator+=(const Vec& r) {*this=*this+r; return *this;}
 Vec& operator-=(const Vec& r) {*this=*this-r; return *this;}
 Vec& operator*=(double r) {*this=*this*r; return *this;}
 Vec& operator/=(double r) {*this=*this/r; return *this;}
 double operator*(const Vec& r) const {return x*r.x+y*r.y;}
 double abs() const {return sqrt(x*x+y*y);}
 double abs_squared() const {return x*x+y*y;}
};

Vec InputVec(){
 double x,y; scanf("%lf%lf",&x,&y);
 return Vec(x,y);
}

void PrintVec(Vec tg){
 printf("%.10f %.10f",tg.x,tg.y);
}

int main() {
 Vec O = InputVec();
 Vec A = InputVec() - O;
 A /= A.abs();
 UL Q; scanf("%u",&Q);
 rep(q,Q){
  Vec X = InputVec() - O;
  Vec Proj = A * (A*X);
  PrintVec(O + Proj + Proj - X);
  printf("\n");
 }
 return 0;
}