
//be naame khodaa
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef complex<ld> PT;
#define F first
#define S second

inline int in(){int x, y; y = scanf("%d", &x); return x; }

const int N = -1;

istream& operator>> (istream& is, complex<ld>& p){
	ld val;
	is >> val;
	p.real(val);
	is >> val;
	p.imag(val);
	return is;
}

PT proj(PT a, PT b, PT c){
	b -= a, c -= a;
	return a + b*real(c/b);
}

PT reflect(PT a, PT b, PT c){
	b -= a, c -= a;
	return a + conj(c/b)*b;
}

void Projection(){
	PT a, b;
	cin >> a >> b;
	cout << setprecision(10) << fixed;
	for (int i = in(); i; i--){
		PT p;
		cin >> p;
		PT img = proj(a, b, p);
		cout << img.real() << ' ' << img.imag() << '\n';
	}
}

void Reflection(){
	PT a, b;
	cin >> a >> b;
	for (int i = in(); i; i--){
		PT p;
		cin >> p;
		PT ref = reflect(a, b, p);
		cout << ref.real() << ' ' << ref.imag() << '\n';
	}
}

int main(){
	cout << setprecision(10) << fixed;
//	Projection();
	Reflection();
	return 0;
}

### Prompt

Generate 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

//be naame khodaa
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef complex<ld> PT;
#define F first
#define S second

inline int in(){int x, y; y = scanf("%d", &x); return x; }

const int N = -1;

istream& operator>> (istream& is, complex<ld>& p){
	ld val;
	is >> val;
	p.real(val);
	is >> val;
	p.imag(val);
	return is;
}

PT proj(PT a, PT b, PT c){
	b -= a, c -= a;
	return a + b*real(c/b);
}

PT reflect(PT a, PT b, PT c){
	b -= a, c -= a;
	return a + conj(c/b)*b;
}

void Projection(){
	PT a, b;
	cin >> a >> b;
	cout << setprecision(10) << fixed;
	for (int i = in(); i; i--){
		PT p;
		cin >> p;
		PT img = proj(a, b, p);
		cout << img.real() << ' ' << img.imag() << '\n';
	}
}

void Reflection(){
	PT a, b;
	cin >> a >> b;
	for (int i = in(); i; i--){
		PT p;
		cin >> p;
		PT ref = reflect(a, b, p);
		cout << ref.real() << ' ' << ref.imag() << '\n';
	}
}

int main(){
	cout << setprecision(10) << fixed;
//	Projection();
	Reflection();
	return 0;
}

```