#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <queue>
#include <numeric>
#include <iomanip>
#include <map>
#include <complex>
using namespace std;
using ll = long long;
#define MOD 1000000007
#define INF 1LL << 59

typedef complex<double> P;

const double EPS = 1e-10;

//二つのスカラーが等しいか
#define EQ(a, b) (abs((a) - (b)) < EPS)
//二つのベクトルが等しいか
#define EQV(a, b) (EQ((a), real(), (b).real()) && EQ((a), imag(), (b).imag()))

double dot(P a, P b)
{
    return real(a) * real(b) + imag(a) * imag(b);
}

double cross(P a, P b)
{
    return real(a) * imag(b) - imag(a) * real(b);
}

ll mod(ll x, ll n)
{
    ll res = 1;
    while (n > 0)
    {
        if ((n & 1) == 1)
        {
            res = (res * x) % MOD;
        }
        x = x * x;
        n >>= 1;
    }
    return res % MOD;
}

int main()
{
    P p1, p2;
    double x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    p1.real(x1);
    p1.imag(y1);
    p2.real(x2);
    p2.imag(y2);
    int q;
    cin >> q;
    while (q--)
    {
        P p;
        double x, y;
        cin >> x >> y;
        p.real(x);
        p.imag(y);

        P ans = p1 + (dot(p - p1, p2 - p1) / (abs(p2 - p1) * abs(p2 - p1))) * (p2 - p1);
        ans.real(2.0 * ans.real() - p.real());
        ans.imag(2.0 * ans.imag() - p.imag());
        cout << fixed << setprecision(12) << ans.real() << " " << ans.imag() << endl;
    }
    return 0;
}

### Prompt

Please provide a Cpp coded solution 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 <vector>
#include <algorithm>
#include <string>
#include <queue>
#include <numeric>
#include <iomanip>
#include <map>
#include <complex>
using namespace std;
using ll = long long;
#define MOD 1000000007
#define INF 1LL << 59

typedef complex<double> P;

const double EPS = 1e-10;

//二つのスカラーが等しいか
#define EQ(a, b) (abs((a) - (b)) < EPS)
//二つのベクトルが等しいか
#define EQV(a, b) (EQ((a), real(), (b).real()) && EQ((a), imag(), (b).imag()))

double dot(P a, P b)
{
    return real(a) * real(b) + imag(a) * imag(b);
}

double cross(P a, P b)
{
    return real(a) * imag(b) - imag(a) * real(b);
}

ll mod(ll x, ll n)
{
    ll res = 1;
    while (n > 0)
    {
        if ((n & 1) == 1)
        {
            res = (res * x) % MOD;
        }
        x = x * x;
        n >>= 1;
    }
    return res % MOD;
}

int main()
{
    P p1, p2;
    double x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    p1.real(x1);
    p1.imag(y1);
    p2.real(x2);
    p2.imag(y2);
    int q;
    cin >> q;
    while (q--)
    {
        P p;
        double x, y;
        cin >> x >> y;
        p.real(x);
        p.imag(y);

        P ans = p1 + (dot(p - p1, p2 - p1) / (abs(p2 - p1) * abs(p2 - p1))) * (p2 - p1);
        ans.real(2.0 * ans.real() - p.real());
        ans.imag(2.0 * ans.imag() - p.imag());
        cout << fixed << setprecision(12) << ans.real() << " " << ans.imag() << endl;
    }
    return 0;
}

```