#define _USE_MATH_DEFINES

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>

using namespace std;

typedef pair<long long int, long long int> P;
long long int INF = 1e18;

int main(){
	
	double x1, x2, y1, y2;
	cin >> x1 >> y1 >> x2 >> y2;
	
	double vx = x2 - x1, vy = y2 - y1;
	double norm = vx * vx + vy * vy;
	
	int q;
	cin >> q;
	
	for(int loop = 0; loop < q; loop++){
		double x, y, vx2, vy2;
		cin >> x >> y;
		vx2 = x - x1;
		vy2 = y - y1;
		double inn = vx * vx2 + vy * vy2;
		double qx = x1 + vx * inn / norm;
		double qy = y1 + vy * inn / norm;
		
		printf("%.10f %.10f\n", 2 * qx - x, 2 * qy - y);
	}
	
	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
#define _USE_MATH_DEFINES

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>

using namespace std;

typedef pair<long long int, long long int> P;
long long int INF = 1e18;

int main(){
	
	double x1, x2, y1, y2;
	cin >> x1 >> y1 >> x2 >> y2;
	
	double vx = x2 - x1, vy = y2 - y1;
	double norm = vx * vx + vy * vy;
	
	int q;
	cin >> q;
	
	for(int loop = 0; loop < q; loop++){
		double x, y, vx2, vy2;
		cin >> x >> y;
		vx2 = x - x1;
		vy2 = y - y1;
		double inn = vx * vx2 + vy * vy2;
		double qx = x1 + vx * inn / norm;
		double qy = y1 + vy * inn / norm;
		
		printf("%.10f %.10f\n", 2 * qx - x, 2 * qy - y);
	}
	
	return 0;
}

```