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 JAVA):

Solution:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        double x_p1 = scan.nextDouble();
        double y_p1 = scan.nextDouble();
        double x_p2 = scan.nextDouble();
        double y_p2 = scan.nextDouble();
        
        double vx = x_p2 - x_p1;
        double vy = y_p2 - y_p1;
        
        int q = scan.nextInt();
        double l = norm(x_p1, y_p1, x_p2, y_p2);
        
        for(int i = 0; i < q; i++) {
            double x = scan.nextDouble();
            double y = scan.nextDouble();
            double ux = x - x_p1;
            double uy = y - y_p1;
            
            double t = dot(ux, uy, vx, vy);
            
            double k = t / l;
            double ax = k * vx + x_p1;
            double ay = k * vy + y_p1;
            
            double bx = 2 * ax - x;
            double by = 2 * ay - y;
            System.out.println(bx + " " + by);
        }
        scan.close();
    }
    
    static double dot(double x1, double y1, double x2, double y2) {
        return x1 * x2 + y1 * y2;
    }
    
    static double norm(double x1, double y1, double x2, double y2) {
        return (x1 - x2) * (x1 - x2) + (y1 -y2) * (y1 - y2);
    }
}