#include<iostream>
#define for0(i, n) for(int i = 0; i < (n); i++)
#define for1(i, n) for(int i = 1; i <= (n);i++)
using namespace std;
int get() { int r; cin >> r; return r; }
signed main() {
	double x1, x2, y1, y2, x3, y3, a, b, c, d;
	cin >> x1 >> y1 >> x2 >> y2;
	a = x1 - x2; b = y1 - y2;
	c = b; d = a;
	if (d < 0)d *= -1;
	else c *= -1;
	int q = get();
	for1(i, q) {
		cin >> x3 >> y3;
		double k1 = -100000, k2 = 100000;
		while (k2 - k1 >= 0.0000000000001) {
			double mid = (k1 + k2) / 2;
			if ((a * (y3 + k1 * d - y1) - b * (x3 + k1 * c - x1)) * ((a * (y3 + mid * d - y1) - b * (x3 + mid * c - x1))) > 0) {
				k1 = mid;
			}
			else k2 = mid;
		}
		double k = k1 + k2;
		printf("%.20f %.20f\n", x3 + k * c, y3 + k * d);
	}
}

### Prompt

Please formulate 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>
#define for0(i, n) for(int i = 0; i < (n); i++)
#define for1(i, n) for(int i = 1; i <= (n);i++)
using namespace std;
int get() { int r; cin >> r; return r; }
signed main() {
	double x1, x2, y1, y2, x3, y3, a, b, c, d;
	cin >> x1 >> y1 >> x2 >> y2;
	a = x1 - x2; b = y1 - y2;
	c = b; d = a;
	if (d < 0)d *= -1;
	else c *= -1;
	int q = get();
	for1(i, q) {
		cin >> x3 >> y3;
		double k1 = -100000, k2 = 100000;
		while (k2 - k1 >= 0.0000000000001) {
			double mid = (k1 + k2) / 2;
			if ((a * (y3 + k1 * d - y1) - b * (x3 + k1 * c - x1)) * ((a * (y3 + mid * d - y1) - b * (x3 + mid * c - x1))) > 0) {
				k1 = mid;
			}
			else k2 = mid;
		}
		double k = k1 + k2;
		printf("%.20f %.20f\n", x3 + k * c, y3 + k * d);
	}
}

```