#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

int main(){
    double x1,y1,x2,y2;
    int q;
    cin >> x1 >> y1 >> x2 >> y2 >> q;
    for(int i=1; i<=q; i++){
        double xp,yp,x,y;
        cin >> xp >> yp;
        double a,b,f;
        if(x1==x2){
            printf("%.12lf %.12lf\n",2*x1-xp,yp);
        }
        else if(y1==y2){
            printf("%.12lf %.12lf\n",xp,2*y1-yp);
        }
        else{
            a=(y2-y1)/(x2-x1);
            b=y1-a*x1;
            f=yp+1.0/a*xp;
            x=(f-b)/(a+1.0/a);
            y=a*x+b;
            x=2*x-xp;
            y=2*y-yp;
            printf("%.12lf %.12lf\n",x,y);
        }
    }
    return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

For given three points p1, p2, p, find the reflection point x of p onto p1p2.

<image>

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

### Response

```cpp
#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

int main(){
    double x1,y1,x2,y2;
    int q;
    cin >> x1 >> y1 >> x2 >> y2 >> q;
    for(int i=1; i<=q; i++){
        double xp,yp,x,y;
        cin >> xp >> yp;
        double a,b,f;
        if(x1==x2){
            printf("%.12lf %.12lf\n",2*x1-xp,yp);
        }
        else if(y1==y2){
            printf("%.12lf %.12lf\n",xp,2*y1-yp);
        }
        else{
            a=(y2-y1)/(x2-x1);
            b=y1-a*x1;
            f=yp+1.0/a*xp;
            x=(f-b)/(a+1.0/a);
            y=a*x+b;
            x=2*x-xp;
            y=2*y-yp;
            printf("%.12lf %.12lf\n",x,y);
        }
    }
    return 0;
}

```