#include <iostream>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <algorithm>
#include <cmath>

#include <cstdio>

using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repc(i, s, n) for (int i = (s); i <= (n); i++)
#define rrep(i, n) for (int i = (n) - 1; i >= 0; i--)
#define rrepc(i, s, n) for (int i = (s); i >= (n); i--)
typedef long long ll;
typedef unsigned long long ull;

struct Point { double x, y; };
typedef Point Vector;
#define EPS (1e-10)
double dot(Vector a, Vector b) {
	return a.x * b.x + a.y * b.y;
}
int main()
{
	int q;
	Point p[3];
	Vector v1, v2;
	double dotv;
	double v1len;
	Point ans, ans2;

	cin >> p[1].x >> p[1].y >> p[2].x >> p[2].y;
	cin >> q;

	v1.x = p[2].x - p[1].x;
	v1.y = p[2].y - p[1].y;
	v1len = sqrt(v1.x * v1.x + v1.y * v1.y);

	rep(i, q) {
		cin >> p[0].x >> p[0].y;

		v2.x = p[0].x - p[1].x;
		v2.y = p[0].y - p[1].y;

		dotv = dot(v1, v2);

		ans.x = p[1].x + v1.x * dotv / v1len / v1len;
		ans.y = p[1].y + v1.y * dotv / v1len / v1len;

		ans2.x = p[0].x + (ans.x - p[0].x) * 2;
		ans2.y = p[0].y + (ans.y - p[0].y) * 2;

		printf("%.10lf %.10lf\n", ans2.x, ans2.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
#include <iostream>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <algorithm>
#include <cmath>

#include <cstdio>

using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repc(i, s, n) for (int i = (s); i <= (n); i++)
#define rrep(i, n) for (int i = (n) - 1; i >= 0; i--)
#define rrepc(i, s, n) for (int i = (s); i >= (n); i--)
typedef long long ll;
typedef unsigned long long ull;

struct Point { double x, y; };
typedef Point Vector;
#define EPS (1e-10)
double dot(Vector a, Vector b) {
	return a.x * b.x + a.y * b.y;
}
int main()
{
	int q;
	Point p[3];
	Vector v1, v2;
	double dotv;
	double v1len;
	Point ans, ans2;

	cin >> p[1].x >> p[1].y >> p[2].x >> p[2].y;
	cin >> q;

	v1.x = p[2].x - p[1].x;
	v1.y = p[2].y - p[1].y;
	v1len = sqrt(v1.x * v1.x + v1.y * v1.y);

	rep(i, q) {
		cin >> p[0].x >> p[0].y;

		v2.x = p[0].x - p[1].x;
		v2.y = p[0].y - p[1].y;

		dotv = dot(v1, v2);

		ans.x = p[1].x + v1.x * dotv / v1len / v1len;
		ans.y = p[1].y + v1.y * dotv / v1len / v1len;

		ans2.x = p[0].x + (ans.x - p[0].x) * 2;
		ans2.y = p[0].y + (ans.y - p[0].y) * 2;

		printf("%.10lf %.10lf\n", ans2.x, ans2.y);
	}

	return 0;
}

```