#include <cmath>
#include <iostream>
#include <iomanip>
using namespace std;

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
	int q; cin >> q;
	long double theta1, theta2;
	theta1 = atan2(y2 - y1, x2 - x1);
	while (q--) {
		int x, y; cin >> x >> y;
		theta2 = atan2(y - y1, x - x1);
		long double dist = hypot(x - x1, y - y1) * cos(theta2 - theta1);
		long double dist_x = dist*cos(theta1), dist_y = dist*sin(theta1);
		cout << fixed << setprecision(11) << 2*x1 + (dist_x - x) + dist_x << " " << 2*y1 + (dist_y - y) + dist_y << endl;
	}
}
### Prompt

Your challenge is to write 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 <cmath>
#include <iostream>
#include <iomanip>
using namespace std;

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
	int q; cin >> q;
	long double theta1, theta2;
	theta1 = atan2(y2 - y1, x2 - x1);
	while (q--) {
		int x, y; cin >> x >> y;
		theta2 = atan2(y - y1, x - x1);
		long double dist = hypot(x - x1, y - y1) * cos(theta2 - theta1);
		long double dist_x = dist*cos(theta1), dist_y = dist*sin(theta1);
		cout << fixed << setprecision(11) << 2*x1 + (dist_x - x) + dist_x << " " << 2*y1 + (dist_y - y) + dist_y << endl;
	}
}
```