Problem: For given three points p1, p2, p, find the reflection point x of p onto p1p2.

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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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using ll = long long;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("Yes")
#define p_no() p("No")

const ll mod = 1e9 + 7;
const ll inf = 1e18;

// 複素数雨セット
typedef complex<double> C;
// 内積
double dot(C a, C b){
    return a.real() * b.real() + a.imag() * b.imag();
}
double round(double a){
    if(abs(a) < 0.00000000001){
        return 0;
    }
    else{
        return a;
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    ll x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;

    C p1(x1, y1);
    C p2(x2, y2);
    C v = p2 - p1;
    C n = v / abs(v);
        
    ll Q; cin >> Q;
    while(Q--){
        ll x, y;
        cin >> x >> y;
        C p(x, y);

        auto v = p - p1;
        auto p_proj = p1 + dot(v, n) * n;
        auto w = p_proj - p;
        auto ans = p_proj + w;

        // cout << round(ans.real()) << ' ' << round(ans.imag()) << endl;
        printf("%.8lf %.8lf\n", ans.real(), ans.imag());
    }
    
    return 0;
}