#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;

typedef struct coo {
    int x;
    int y;
};

int main() {
    coo p1, p2;
    cin >> p1.x >> p1.y >> p2.x >> p2.y;
    int q;
    cin >> q;
    coo* p = new coo[q];
    for (int i = 0; i < q; i++) {
        cin >> p[i].x >> p[i].y;
        int naiseki = (p[i].x - p1.x) * (p2.x - p1.x) + (p[i].y - p1.y) * (p2.y - p1.y);
        double leng = (double)naiseki / (double)((p2.x - p1.x) * (p2.x - p1.x) + (p2.y - p1.y) * (p2.y - p1.y));
        printf("%6.10f %6.10f\n", 2*(p1.x + (p2.x - p1.x) * leng)-p[i].x, 2*(p1.y + (p2.y - p1.y) * leng)-p[i].y);
    }
    delete[] p;
    return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;

typedef struct coo {
    int x;
    int y;
};

int main() {
    coo p1, p2;
    cin >> p1.x >> p1.y >> p2.x >> p2.y;
    int q;
    cin >> q;
    coo* p = new coo[q];
    for (int i = 0; i < q; i++) {
        cin >> p[i].x >> p[i].y;
        int naiseki = (p[i].x - p1.x) * (p2.x - p1.x) + (p[i].y - p1.y) * (p2.y - p1.y);
        double leng = (double)naiseki / (double)((p2.x - p1.x) * (p2.x - p1.x) + (p2.y - p1.y) * (p2.y - p1.y));
        printf("%6.10f %6.10f\n", 2*(p1.x + (p2.x - p1.x) * leng)-p[i].x, 2*(p1.y + (p2.y - p1.y) * leng)-p[i].y);
    }
    delete[] p;
    return 0;
}

```