#include "bits/stdc++.h"

#define REP(i,num) for(int i=0;i<(num);++i)
#define ALL(c) c.begin(),c.end()
#define PRINTALL(c) for(auto& x:c){cout<<x<<' ';}cout<<endl;
#define PAIRCOMP(c,comp) [](const pair<ll,ll>& lhs,const pair<ll,ll>& rhs){return lhs.c comp rhs.c;}

using namespace std;
using ll = long long;

constexpr ll atcoder_mod = 1e9+7;

template<typename T=int>
T in(){T x; cin >> x; return (x);}
template<typename T=int,typename C=vector<T>>
C vecin(int N){C x(N);REP(i,N){x[i]=in<T>();}return move(x);}

void vout(){cout << endl;}
template<typename Head,typename... Tail>
void vout(Head&& h,Tail&&... t){cout << ' ' << h;vout(forward<Tail>(t)...);}
void out(){cout << endl;}
template<typename Head,typename... Tail>
void out(Head&& h,Tail&&... t){cout << h;vout(forward<Tail>(t)...);}

int main(){
	//cin.tie(0);
	//ios::sync_with_stdio(false);

	auto x1=in(),y1=in(),x2=in(),y2=in();
	double dx = x2-x1,dy = y2-y1;
	double norm = hypot(dx,dy);
	dx/=norm,dy/=norm;
	auto Q=in();
	REP(i,Q){
		int x3=in(),y3=in();
		double proj = dx*(x3-x1)+dy*(y3-y1);
		printf("%.10lf %.10lf\n",2*(proj*dx+x1)-x3,2*(proj*dy+y1)-y3);
	}

	return 0;
}


### Prompt

Create a solution in Cpp for 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 "bits/stdc++.h"

#define REP(i,num) for(int i=0;i<(num);++i)
#define ALL(c) c.begin(),c.end()
#define PRINTALL(c) for(auto& x:c){cout<<x<<' ';}cout<<endl;
#define PAIRCOMP(c,comp) [](const pair<ll,ll>& lhs,const pair<ll,ll>& rhs){return lhs.c comp rhs.c;}

using namespace std;
using ll = long long;

constexpr ll atcoder_mod = 1e9+7;

template<typename T=int>
T in(){T x; cin >> x; return (x);}
template<typename T=int,typename C=vector<T>>
C vecin(int N){C x(N);REP(i,N){x[i]=in<T>();}return move(x);}

void vout(){cout << endl;}
template<typename Head,typename... Tail>
void vout(Head&& h,Tail&&... t){cout << ' ' << h;vout(forward<Tail>(t)...);}
void out(){cout << endl;}
template<typename Head,typename... Tail>
void out(Head&& h,Tail&&... t){cout << h;vout(forward<Tail>(t)...);}

int main(){
	//cin.tie(0);
	//ios::sync_with_stdio(false);

	auto x1=in(),y1=in(),x2=in(),y2=in();
	double dx = x2-x1,dy = y2-y1;
	double norm = hypot(dx,dy);
	dx/=norm,dy/=norm;
	auto Q=in();
	REP(i,Q){
		int x3=in(),y3=in();
		double proj = dx*(x3-x1)+dy*(y3-y1);
		printf("%.10lf %.10lf\n",2*(proj*dx+x1)-x3,2*(proj*dy+y1)-y3);
	}

	return 0;
}


```