#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <unordered_map>
#include <cstdlib>
using namespace std;
#define Ma_PI 3.141592653589793
#define eps 0.00001
#define LONG_INF 3000000000000000000
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007
#define MOD 998244353
#define REP(i,n) for(long long i = 0;i < n;++i)    
#define seg_size 524288
#include <complex>
typedef complex<double> P;
double dot(P a, P b) {
	return (a.real() * b.real() + a.imag() * b.imag());
}
double cross(P a, P b) {
	return (a.real() * b.imag() - a.imag() * b.real());
}
int main() {
	double a, b, c, d;
	cin >> a >> b >> c >> d;
	P line = P(c - a, d - b);
	int query;
	cin >> query;
	REP(i, query) {
		double e, f;
		cin >> e >> f;
		P hypo = P(e - a, f - b);
		P hoge = P(a, b) + line * dot(line, hypo) / pow(abs(line), 2);
		P final_digit = hoge - P(e, f);
		P ans = P(e, f) + final_digit * 2.0;
		cout << fixed << setprecision(10) << ans.real() << " " << ans.imag() << endl;
	}
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <unordered_map>
#include <cstdlib>
using namespace std;
#define Ma_PI 3.141592653589793
#define eps 0.00001
#define LONG_INF 3000000000000000000
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007
#define MOD 998244353
#define REP(i,n) for(long long i = 0;i < n;++i)    
#define seg_size 524288
#include <complex>
typedef complex<double> P;
double dot(P a, P b) {
	return (a.real() * b.real() + a.imag() * b.imag());
}
double cross(P a, P b) {
	return (a.real() * b.imag() - a.imag() * b.real());
}
int main() {
	double a, b, c, d;
	cin >> a >> b >> c >> d;
	P line = P(c - a, d - b);
	int query;
	cin >> query;
	REP(i, query) {
		double e, f;
		cin >> e >> f;
		P hypo = P(e - a, f - b);
		P hoge = P(a, b) + line * dot(line, hypo) / pow(abs(line), 2);
		P final_digit = hoge - P(e, f);
		P ans = P(e, f) + final_digit * 2.0;
		cout << fixed << setprecision(10) << ans.real() << " " << ans.imag() << endl;
	}
	return 0;
}

```