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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll INF = 1e18;
const int BIG = 1e9 + 555;

struct Point {
	double x, y;
	Point(){};
	Point(double _x, double _y) : x(_x), y(_y) {};
};

Point operator + (const Point &p, const Point &q){
	return Point(p.x + q.x, p.y + q.y);
}

Point operator - (const Point &p, const Point &q){
	return Point(p.x - q.x, p.y - q.y);
}

double operator * (const Point &p, const Point &q){
	return p.x * q.x + p.y * q.y;
}

double dist(const Point &p, const Point &q){
	return sqrt((q - p) * (q - p));
}

Point operator * (const Point &p, double k){
	return Point(p.x * k, p.y * k);
}

Point projection(const Point &A, const Point &p, const Point &q){
	Point pq = q - p;
	Point pA = A - p;
	
	double K = (pq * pA) / (pq * pq);

	return p + pq * K;
}

Point reflection(const Point &A, const Point &p, const Point &q){
	Point m = projection(A, p, q);
	///2 * m.x = (A.x + B.x)

	return Point(2 * m.x - A.x, 2 * m.y - A.y);
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cout << setprecision(10) << fixed;

	Point p1, p2;
	cin >> p1.x >> p1.y >> p2.x >> p2.y;

	int q;	cin >> q;
	while(q--){
		Point p;	cin >> p.x >> p.y;
		Point ref = reflection(p, p1, p2);

		cout << ref.x << ' ' << ref.y << '\n';
	}

	return 0;
}