#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<stack>
#include<climits>
#include<cstring>
#include<queue>
using namespace std;

vector<double> ans(2);

void reflection(double x1, double y1, double x2, double y2, double x, double y){
	double keisu = ((x-x1)*(x2-x1)+(y-y1)*(y2-y1))/((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
	double s1=2*(keisu*(x2-x1)+x1)-x;
	double s2=2*(keisu*(y2-y1)+y1)-y;
	ans[0]=s1;
	ans[1]=s2;
}

int main(){
	double x1,y1,x2,y2,q,x,y;
	cin >> x1 >> y1 >> x2 >> y2 >> q ;
	for(int i=0;i<q;i++){
		cin >> x >> y;
		reflection(x1,y1,x2,y2,x,y);
		printf("%.10lf %.10lf \n",ans[0],ans[1]);
	}
	return 0;
}
### Prompt

Your task is to create a cpp solution 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<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<stack>
#include<climits>
#include<cstring>
#include<queue>
using namespace std;

vector<double> ans(2);

void reflection(double x1, double y1, double x2, double y2, double x, double y){
	double keisu = ((x-x1)*(x2-x1)+(y-y1)*(y2-y1))/((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
	double s1=2*(keisu*(x2-x1)+x1)-x;
	double s2=2*(keisu*(y2-y1)+y1)-y;
	ans[0]=s1;
	ans[1]=s2;
}

int main(){
	double x1,y1,x2,y2,q,x,y;
	cin >> x1 >> y1 >> x2 >> y2 >> q ;
	for(int i=0;i<q;i++){
		cin >> x >> y;
		reflection(x1,y1,x2,y2,x,y);
		printf("%.10lf %.10lf \n",ans[0],ans[1]);
	}
	return 0;
}
```