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 <iostream>
#include <vector>
#include <utility>
#include <cstdio>
using namespace std;

int main(){
  double xp1,yp1,xp2,yp2;
  cin >> xp1 >> yp1 >> xp2 >> yp2;
  pair<double,double> p1 = make_pair(xp1,yp1);
  pair<double,double> p2 = make_pair(xp2,yp2);

  pair<double,double> a = make_pair(p2.first-p1.first,p2.second-p1.second);

  int q;
  cin >> q;
  vector<pair<double,double>> query;
  for(int i = 0;i < q;i++){
    double xp,yp;
    cin >> xp >> yp;
    pair<double,double> p = make_pair(xp,yp);
    pair<double,double> b = make_pair(p.first-p1.first,p.second-p1.second);

    //同様に正射影を用いればよい.
    double dot_ab = a.first * b.first + a.second * b.second;
    double dot_aa = a.first * a.first + a.second * a.second;
    pair<double,double> prj = make_pair(a.first*dot_ab/dot_aa,a.second*dot_ab/dot_aa);

    //prj=(p1->p(=b) + p1->x)/2
    pair<double,double> p1x = make_pair(2*prj.first-b.first,2*prj.second-b.second);
    pair<double,double> ans = make_pair(p1.first+p1x.first,p1.second+p1x.second);

    query.push_back(ans);
  }

  for(int i = 0;i < q;i++){
    printf("%.14lf %.14lf\n",query[i].first,query[i].second);
  }
}