#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pairInt;
#define FOR(i, n) for (int i = 0; i < int(n); i++)
#define FOR1(i, m, n) for (int i = int(m); i < int(n); i++)
#define MAX(a, b) ((a) >= (b) ? (a) : (b))
#define MIN(a, b) ((a) <= (b) ? (a) : (b))

int main(int argc, char *argv[]) {
  int q;
  array<int, 2> p1, p2, p, v1, v2;

  scanf("%d%d%d%d%d", &p1[0], &p1[1], &p2[0], &p2[1], &q);
  v1[0] = p2[0] - p1[0];
  v1[1] = p2[1] - p1[1];

  FOR(i, q) {
    scanf("%d%d", &p[0], &p[1]);
    v2[0] = p[0] - p1[0];
    v2[1] = p[1] - p1[1];

    int ip1 = inner_product(v1.begin(), v1.end(), v1.begin(), 0);
    int ip2 = inner_product(v2.begin(), v2.end(), v1.begin(), 0);
    array<double, 2> x = {
      p1[0] + (double)ip2 / ip1 * (p2[0] - p1[0]),
      p1[1] + (double)ip2 / ip1 * (p2[1] - p1[1])
    };

    printf("%.10lf %.10lf\n", 2 * x[0] - p[0], 2 * x[1] - p[1]);
  }

  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 <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pairInt;
#define FOR(i, n) for (int i = 0; i < int(n); i++)
#define FOR1(i, m, n) for (int i = int(m); i < int(n); i++)
#define MAX(a, b) ((a) >= (b) ? (a) : (b))
#define MIN(a, b) ((a) <= (b) ? (a) : (b))

int main(int argc, char *argv[]) {
  int q;
  array<int, 2> p1, p2, p, v1, v2;

  scanf("%d%d%d%d%d", &p1[0], &p1[1], &p2[0], &p2[1], &q);
  v1[0] = p2[0] - p1[0];
  v1[1] = p2[1] - p1[1];

  FOR(i, q) {
    scanf("%d%d", &p[0], &p[1]);
    v2[0] = p[0] - p1[0];
    v2[1] = p[1] - p1[1];

    int ip1 = inner_product(v1.begin(), v1.end(), v1.begin(), 0);
    int ip2 = inner_product(v2.begin(), v2.end(), v1.begin(), 0);
    array<double, 2> x = {
      p1[0] + (double)ip2 / ip1 * (p2[0] - p1[0]),
      p1[1] + (double)ip2 / ip1 * (p2[1] - p1[1])
    };

    printf("%.10lf %.10lf\n", 2 * x[0] - p[0], 2 * x[1] - p[1]);
  }

  return 0;
}
```