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 <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>

#define mp       make_pair
#define pb       push_back
#define all(x)   (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define repi(i,a,b) for(int i = (a); i < (b); i++)

using namespace std;

typedef    long long          ll;
typedef    unsigned long long ull;
typedef    vector<bool>       vb;
typedef    vector<int>        vi;
typedef    vector<vb>         vvb;
typedef    vector<vi>         vvi;
typedef    pair<int,int>      pii;

const int INF=1<<29;
const double EPS=1e-9;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};
//射影
//基本的な考えは(x1,y1)と(x2,y2)の線分lに対して点Qに(xq,rq)のlを対称軸として点Rを(xr,yr)と点Qの中点を取れば点Qから線lへの垂線との交点が求まる
int main(){
	double x1,y1,x2,y2,xq,yq;
	double m,x,y,c;
	scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
	int q;
	cin>>q;
	for(int i =0;i < q;i++){
		scanf("%lf %lf",&xq,&yq);
		if(x2==x1){
			xq *=-1;
			printf("%.9f %.9f\n",xq,yq);
		}else{
			m = (y2-y1)/(x2-x1);
			c = y2-m*x2;
			x = (2*m*yq-xq*(m*m-1)-2*m*c)/(m*m+1);
			y = (2*m*xq+yq*(m*m-1)+2*c)/(m*m+1);
			printf("%.9f %.9f\n", x, y);
		}
	}
	return 0;
}