#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<utility>
#include<vector>
#include<cmath>
#include<cstdio>
#include<complex>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define it ::iterator
#define all(in) in.begin(),in.end()
const double PI=acos(-1);
const double ESP=1e-10;
using namespace std;
typedef complex<double> P;
P pin(){
	double x,y;
	cin>>x>>y;
	P p(x,y);
	return p;
}
void PIN(P* a,int i){
	rep(i,2)a[i]=pin();
}
double dot(P a,P b){
	return real(conj(a)*b);
}
double cross(P a,P b){
	return imag(conj(a)*b);
}
int main(){
	P a[2];
	PIN(a,2);
	int n;
	cin>>n;
	rep(i,n){
		P p2=pin();
		P p=a[1]-a[0];
		P t=a[0]+dot(p2-a[0],p)*p/norm(p);
		P ans=(t-p2)*2.0+p2;
		printf("%.9f %.9f\n",real(ans),imag(ans));
	}
		


}
### 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<string>
#include<algorithm>
#include<map>
#include<set>
#include<utility>
#include<vector>
#include<cmath>
#include<cstdio>
#include<complex>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define it ::iterator
#define all(in) in.begin(),in.end()
const double PI=acos(-1);
const double ESP=1e-10;
using namespace std;
typedef complex<double> P;
P pin(){
	double x,y;
	cin>>x>>y;
	P p(x,y);
	return p;
}
void PIN(P* a,int i){
	rep(i,2)a[i]=pin();
}
double dot(P a,P b){
	return real(conj(a)*b);
}
double cross(P a,P b){
	return imag(conj(a)*b);
}
int main(){
	P a[2];
	PIN(a,2);
	int n;
	cin>>n;
	rep(i,n){
		P p2=pin();
		P p=a[1]-a[0];
		P t=a[0]+dot(p2-a[0],p)*p/norm(p);
		P ans=(t-p2)*2.0+p2;
		printf("%.9f %.9f\n",real(ans),imag(ans));
	}
		


}
```