#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define cinf(n,x) for(int i=0;i<(n);i++)cin>>x[i];
#define ft first
#define sc second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(v) (v).begin(),(v).end()
#define mod 1000000007
using namespace std;
typedef long long ll;
template<class T> using V=vector<T>;
using Graph = vector<vector<int>>;
using P=pair<ll,ll>;
typedef unsigned long long ull;
typedef long double ldouble;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
//V,P(大文字)使用不可
//乗算の際にオーバーフローに注意せよ！！（適切にmodをとれ）
//制約をよく読め！

const ll INF=1e18;

signed main(){
    V<double> x(2),y(2);
    rep(i,2) cin>>x[i]>>y[i];
    int q;
    cin>>q;
    while(q--){
        double a,b;
        cin>>a>>b;
        double t=((x[0]-x[1])*(x[0]-a)+(y[0]-y[1])*(y[0]-b))/(pow(x[1]-x[0],2)+pow(y[1]-y[0],2));
        double xx=x[0]+t*(x[1]-x[0]);
        double yy=y[0]+t*(y[1]-y[0]);
        double px=xx-a,py=yy-b;
        cout<<fixed<<setprecision(10)<<a+2*px<<' '<<b+2*py<<endl;
    }
}

### Prompt

Your challenge is to write 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
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define cinf(n,x) for(int i=0;i<(n);i++)cin>>x[i];
#define ft first
#define sc second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(v) (v).begin(),(v).end()
#define mod 1000000007
using namespace std;
typedef long long ll;
template<class T> using V=vector<T>;
using Graph = vector<vector<int>>;
using P=pair<ll,ll>;
typedef unsigned long long ull;
typedef long double ldouble;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
//V,P(大文字)使用不可
//乗算の際にオーバーフローに注意せよ！！（適切にmodをとれ）
//制約をよく読め！

const ll INF=1e18;

signed main(){
    V<double> x(2),y(2);
    rep(i,2) cin>>x[i]>>y[i];
    int q;
    cin>>q;
    while(q--){
        double a,b;
        cin>>a>>b;
        double t=((x[0]-x[1])*(x[0]-a)+(y[0]-y[1])*(y[0]-b))/(pow(x[1]-x[0],2)+pow(y[1]-y[0],2));
        double xx=x[0]+t*(x[1]-x[0]);
        double yy=y[0]+t*(y[1]-y[0]);
        double px=xx-a,py=yy-b;
        cout<<fixed<<setprecision(10)<<a+2*px<<' '<<b+2*py<<endl;
    }
}

```