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>

#define NIL -2
#define INF 1000000001
#define EPS 1e-10
#define equals(a,b) (fabs((a) - (b)) < EPS)

using namespace std;
class Point{
    public:
    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 k){ return Point(x*k,y*k);      }
    Point operator / (double k){ return Point(x/k,y/k);      }

    double norm(){ return x*x + y*y;}
    double abs(){ return sqrt(norm());}
    double dot(Point p){
        return x*p.x + y*p.y;
    }

    bool operator < (Point &p){
        return x != p.x ? x < p.x : y < p.y;
    }
    bool operator == (Point &p){
        return fabs(x-p.x) < EPS && fabs(y-p.y) < EPS;
    }
  
};
typedef Point Vector;

int main(){
    double x1,y1,x2,y2;
    double x,y;
    scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
    Point p1(x1,y1);
    Point p2(x2,y2);
    Vector p12 = p2-p1;
    p12 = p12 / p12.abs();

    int q;
    scanf("%d",&q);
    for(int i = 0;i < q;i++){
        scanf("%lf %lf",&x,&y);
        Point p(x,y);
        p = p - p1;
        Point ans;
        double inner_prod = p12.dot(p);
        ans = p + (p12 * inner_prod - p)*2 +p1;
        printf("%.9lf %.9lf\n",ans.x,ans.y);
    }
    return 0;
}