// #include {{{
#include <iostream>
#include <cassert>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <bitset>
#include <vector>
#include <complex>
#include <algorithm>
using namespace std;
// }}}
// #define {{{
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define per(i,a,b) for(int i=(b)-1;i>=(a);--i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
// }}}

typedef db T;
struct P{
    T x,y;
    P(){} P(T x,T y):x(x),y(y){}
    P operator - (const P&b) const {return P(x-b.x,y-b.y);}
    P operator + (const P&b) const {return P(x+b.x,y+b.y);}
    P operator * (const T&k) const {return P(x*k,y*k);}
    T operator * (const P&b) const {return x*b.x+y*b.y;}
    void read() {scanf("%lf%lf",&x,&y);}
};
T norm(P a){return a.x*a.x+a.y*a.y;}
P proj(P p,P a,P b){return (b-a)*((p-a)*(b-a)/norm(b-a))+a;}
P reflect(P p,P a,P b){return proj(p,a,b)*2-p;}

int main(){
    P p,a,b;
    a.read();b.read();
    int Q;scanf("%d",&Q);
    rep(i,0,Q){
        p.read();
        P q=reflect(p,a,b);
        printf("%.10f %.10f\n",q.x,q.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 {{{
#include <iostream>
#include <cassert>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <bitset>
#include <vector>
#include <complex>
#include <algorithm>
using namespace std;
// }}}
// #define {{{
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define per(i,a,b) for(int i=(b)-1;i>=(a);--i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
// }}}

typedef db T;
struct P{
    T x,y;
    P(){} P(T x,T y):x(x),y(y){}
    P operator - (const P&b) const {return P(x-b.x,y-b.y);}
    P operator + (const P&b) const {return P(x+b.x,y+b.y);}
    P operator * (const T&k) const {return P(x*k,y*k);}
    T operator * (const P&b) const {return x*b.x+y*b.y;}
    void read() {scanf("%lf%lf",&x,&y);}
};
T norm(P a){return a.x*a.x+a.y*a.y;}
P proj(P p,P a,P b){return (b-a)*((p-a)*(b-a)/norm(b-a))+a;}
P reflect(P p,P a,P b){return proj(p,a,b)*2-p;}

int main(){
    P p,a,b;
    a.read();b.read();
    int Q;scanf("%d",&Q);
    rep(i,0,Q){
        p.read();
        P q=reflect(p,a,b);
        printf("%.10f %.10f\n",q.x,q.y);
    }
    return 0;
}
```