#include<iostream>
#include<iomanip>
using namespace std;

int main()
{
    double xp1, yp1, xp2, yp2, xp, yp;
    int q;

    cin >> xp1 >> yp1 >> xp2 >> yp2;
    cin >> q;

    //直線p1p2の方程式作り、p1p2の傾きとpxの傾きの積が-1になることを利用して、直線pxの方程式を作る。2つの直線の方程式から交点を求め完了。
    for( int i = 0; i < q; i++ )
    {
        cin >> xp >> yp;

        double a, b, mx, my, x, y;
        a = xp2 - xp1;
        b = yp2 - yp1;
        
        //p1とp2が縦に並んだ場合
        if( a == 0 )
            cout << fixed << setprecision(10) << xp1 + ( xp1 - xp ) << " " << yp << endl;
        //p1とp2が横に並んだ場合
        else if( b == 0 )
            cout << fixed << setprecision(10) << xp << " " << yp1 + ( yp1 - yp ) << endl;
        else
        {
            mx = ( yp - yp1 + a * xp / b + b * xp1 / a ) / ( b / a + a / b );
            my = -1 * a * mx / b + yp + a * xp / b;
            x = mx + ( mx - xp );
            y = my + ( my - yp );
            cout << fixed << setprecision(10) << x << " " << y << endl;
        }
    }
    return 0;
}

### Prompt

Please create a solution in CPP to 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<iomanip>
using namespace std;

int main()
{
    double xp1, yp1, xp2, yp2, xp, yp;
    int q;

    cin >> xp1 >> yp1 >> xp2 >> yp2;
    cin >> q;

    //直線p1p2の方程式作り、p1p2の傾きとpxの傾きの積が-1になることを利用して、直線pxの方程式を作る。2つの直線の方程式から交点を求め完了。
    for( int i = 0; i < q; i++ )
    {
        cin >> xp >> yp;

        double a, b, mx, my, x, y;
        a = xp2 - xp1;
        b = yp2 - yp1;
        
        //p1とp2が縦に並んだ場合
        if( a == 0 )
            cout << fixed << setprecision(10) << xp1 + ( xp1 - xp ) << " " << yp << endl;
        //p1とp2が横に並んだ場合
        else if( b == 0 )
            cout << fixed << setprecision(10) << xp << " " << yp1 + ( yp1 - yp ) << endl;
        else
        {
            mx = ( yp - yp1 + a * xp / b + b * xp1 / a ) / ( b / a + a / b );
            my = -1 * a * mx / b + yp + a * xp / b;
            x = mx + ( mx - xp );
            y = my + ( my - yp );
            cout << fixed << setprecision(10) << x << " " << y << endl;
        }
    }
    return 0;
}

```