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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <complex>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <functional>
using namespace std;

#define Rep(b, e, i) for(int i = b; i <= e; i++)
#define Repr(e, b, i) for(int i = e; i >= b; i--)
#define rep(n, i) Rep(0, n-1, i)
#define repr(n, i) Repr(n-1, 0, i)
#define all(v) (v).begin(), (v).end()
#define pb(v) push_back(v)
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define bitcnt(x) __builtin_popcount(x)
#define fst first
#define snd second
#define Pqaz(T) priority_queue<T,vector<T>,greater<T>>
#define Pqza(T) priority_queue<T>
#define put(x) cout << x;
#define puts(x) cout << x << ' ';
#define putln(x) cout << x << endl;
#define ENJYU std::ios::sync_with_stdio(false);std::cin.tie(0);

typedef complex<double> comp;

void solve(void){
	int x1, y1, x2, y2, Q;
	cin >> x1 >> y1 >> x2 >> y2 >> Q;

	comp o(x1, y1), p(x2-x1, y2-y1), r;

	r = polar(1.0, arg(p));

	while(Q--)
	{
		int x3, y3;
		cin >> x3 >> y3;

		comp q(x3-x1, y3-y1);

		q = conj(q / r) * r + o;

		printf("%.12lf %.12lf\n", q.real(), q.imag());
	}


}
int main(void){
	solve();
	//cout << "yui(*-v・)yui" << endl;
	return 0;
}