#include<bits/stdc++.h>
#define _ ios_base::sync_with_stdio(0);cin.tie(0);
#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;
typedef pair<long double, long double> p;
#define X first
#define Y second

p transition(p p0, p p1, p p2){
  p t, t0;
  double theta = atan2(p1.Y-p0.Y,p1.X-p0.X);
  double s = sin(theta), c = cos(theta);
  t0.X = c*(p2.X-p0.X)+s*(p2.Y-p0.Y);
  t0.Y = -s*(p2.X-p0.X)+c*(p2.Y-p0.Y);
  t.X = c*t0.X+s*t0.Y+p0.X;
  t.Y = s*t0.X-c*t0.Y+p0.Y;
  return t;
}

int main(){ _;
  p p0,p1,p2,t;
  int q;
  cin>>p0.X>>p0.Y>>p1.X>>p1.Y>>q;
  REP(i,q){
    cin>>p2.X>>p2.Y;
    t = transition(p0,p1,p2);
    printf("%.10Lf %.10Lf\n", t.X, t.Y);
  }
}
### 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 _ ios_base::sync_with_stdio(0);cin.tie(0);
#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;
typedef pair<long double, long double> p;
#define X first
#define Y second

p transition(p p0, p p1, p p2){
  p t, t0;
  double theta = atan2(p1.Y-p0.Y,p1.X-p0.X);
  double s = sin(theta), c = cos(theta);
  t0.X = c*(p2.X-p0.X)+s*(p2.Y-p0.Y);
  t0.Y = -s*(p2.X-p0.X)+c*(p2.Y-p0.Y);
  t.X = c*t0.X+s*t0.Y+p0.X;
  t.Y = s*t0.X-c*t0.Y+p0.Y;
  return t;
}

int main(){ _;
  p p0,p1,p2,t;
  int q;
  cin>>p0.X>>p0.Y>>p1.X>>p1.Y>>q;
  REP(i,q){
    cin>>p2.X>>p2.Y;
    t = transition(p0,p1,p2);
    printf("%.10Lf %.10Lf\n", t.X, t.Y);
  }
}
```