#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

#define EPS (1e-10)
#define equals (fabs((a)-(b)<EPS)
int n;

struct Point
{
    double x, y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
    Point operator +(Point a)
    {
        return Point(x+a.x,y+a.y);
    }
    Point operator -(Point a)
    {
        return Point(x-a.x,y-a.y);
    }
    Point operator *(double a)
    {
        return Point(x*a,y*a);
    }

    double norm()
    {
        return x*x+y*y;
    }
};
Point p1,p2,p;
double cross(Point a,Point b)
{
    return a.x*b.y-b.x*a.y;
}
double dot(Point a,Point b)
{
    return a.x*b.x+a.y*b.y;
}
void solve()
{
    Point a=p2-p1;
    Point b=p-p1;
    double r=dot(a,b)/a.norm();
    Point q=p1+a*r;
    Point s=p+(q-p)*2;
    printf("%.8lf %.8lf\n",s.x,s.y);
}

int main()
{
//    freopen("in.txt","r",stdin);
    scanf("%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y);
    scanf("%d",&n);
    while(n--)
    {
        scanf("%lf%lf",&p.x,&p.y);
        solve();
    }
    return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

#define EPS (1e-10)
#define equals (fabs((a)-(b)<EPS)
int n;

struct Point
{
    double x, y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
    Point operator +(Point a)
    {
        return Point(x+a.x,y+a.y);
    }
    Point operator -(Point a)
    {
        return Point(x-a.x,y-a.y);
    }
    Point operator *(double a)
    {
        return Point(x*a,y*a);
    }

    double norm()
    {
        return x*x+y*y;
    }
};
Point p1,p2,p;
double cross(Point a,Point b)
{
    return a.x*b.y-b.x*a.y;
}
double dot(Point a,Point b)
{
    return a.x*b.x+a.y*b.y;
}
void solve()
{
    Point a=p2-p1;
    Point b=p-p1;
    double r=dot(a,b)/a.norm();
    Point q=p1+a*r;
    Point s=p+(q-p)*2;
    printf("%.8lf %.8lf\n",s.x,s.y);
}

int main()
{
//    freopen("in.txt","r",stdin);
    scanf("%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y);
    scanf("%d",&n);
    while(n--)
    {
        scanf("%lf%lf",&p.x,&p.y);
        solve();
    }
    return 0;
}
```