Problem:
Score : 800 points

Problem Statement

There are N lines in the xy-plane. The i-th line is represented by A_ix+B_iy=C_i.
Any two lines among the N+2 lines, the above N lines plus the x-axis and y-axis, cross each other at exactly one point.

For each pair 1 \leq i < j \leq N, there is a car at the cross point of the i-th and j-th lines.
Even where three or more lines intersect at a point, a car is individually placed for each pair of lines.
That is, there will be k(k-1)/2 cars placed at the intersection of k lines.

Those cars are already very old, and can only be moved parallel to the x-axis or y-axis.

Takahashi will hold an exhibition of antique cars at a place on the xy-plane.
In order to avoid damaging the half-broken cars too much, he will select the place of the exhibition so that the total distance covered will be minimized when all the cars are moved to the place.
If such a place is not uniquely determined, among the places that satisfy the condition above, the place with the minimum x-coordinate will be selected.
If the place is still not uniquely determined, among the places that satisfy the two conditions above, the place with the minimum y-coordinate will be selected.

Find the place of the exhibition that will be selected.

Constraints

2 \leq N \leq 4 × 10^4

1 \leq |A_i|,|B_i| \leq 10^4(1 \leq i \leq N)

0 \leq |C_i| \leq 10^4(1 \leq i \leq N)

No two given lines are parallel.

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
A_1 B_1 C_1
:
A_N B_N C_N

Outputs

Print the x-coordinate and y-coordinate of the place of the exhibition that will be selected, in this order, with a space in between. The output will be judged as correct when the absolute or relative error is at most 10^{-9}.

Sample Input 1

3
1 1 1
2 -1 2
-1 2 2

Sample Output 1

1.000000000000000 1.000000000000000

There is a car at each place shown by a blue circle in the figure. The place to be selected is shown by a purple circle.

Sample Input 2

4
1 1 2
1 -1 0
3 -1 -2
1 -3 4

Sample Output 2

-1.000000000000000 -1.000000000000000

Sample Input 3

7
1 7 8
-2 4 9
3 -8 -5
9 2 -14
6 7 5
-8 -9 3
3 8 10

Sample Output 3

-1.722222222222222 1.325000000000000

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//#include <bits/stdc++.h>

#include <iostream>
#include <algorithm>

#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <cstring>
#include <string>
#include <utility>
#include <array>
#include <complex>
#include <valarray>

#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>


using namespace std;
#define int long long

typedef long long ll;
typedef unsigned long long ull;
//typedef unsigned __int128 HASH;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pullull;
typedef pair<ll,int> plli;
typedef pair<long double, int> pdbi;
typedef pair<int,pii> pipii;
typedef pair<ll,pll> plpll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<pii> vpii;
typedef vector<vector<int>> mat;

#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n);i>0;i--)
#define rrep2(i,a,b) for (int i=(a);i>b;i--)
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()

const ll hmod1 = 999999937;
const ll hmod2 = 1000000000 + 9;
const int INF = 1<<30;
const ll INFLL = 1LL<<62;
const long double EPS = 1e-12;
const ll mod = 1000000000 + 7;
const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
const int dy8[8] = {0, 1, -1, 1, -1, 0, 1, -1};
const long double pi = 3.141592653589793;

#define addm(X, Y) (X) = ((X) + ((Y) % mod) + mod) % mod
#define inside(y, x, h, w) (0 <= (y) && (y) < (h) && 0 <= (x) && (x) < (w)) ? true : false

//debug
#define DEBUG

#define DUMPOUT cout

#ifdef DEBUG
#define dump(...) DUMPOUT<<#__VA_ARGS__<<" :["<<__FUNCTION__<<":"<<__LINE__<<"]"<<endl; DUMPOUT<<"    "; dump_func(__VA_ARGS__)
#else
#define dump(...)
#endif

void dump_func() {DUMPOUT << endl;};

template <class Head, class... Tail> void dump_func(Head&& head, Tail&&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) == 0) DUMPOUT << " ";
    else DUMPOUT << ", ";
    dump_func(std::move(tail)...);
}

//ostream
template<typename T> ostream& operator << (ostream& os, vector<T>& vec) {
    os << "["; for (int i = 0; i<vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : ", "); os << "]";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, pair<T, U>& pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, map<T, U>& map_var) {
    os << "[";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")"; itr++;  if(itr != map_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

template<typename T> ostream& operator << (ostream& os, set<T>& set_var) {
    os << "[";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr; ++itr; if(itr != set_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

int n;
vector<pair<long double, int>> fx, fy;
int posx[40000 + 4], posy[40000 + 4];
long double a[40000 + 4], b[40000 + 4], c[40000 + 4];

struct BIT {
    int size;
    vector<int> node;

    BIT(int n, long long init_num = 0LL) {
        size = n;
        node.resize(size + 1, init_num);
    }

    void add(int i, long long x) {
        while (i <= size) {
            node[i] += x;
            i += i & -i;
        }
    }

    long long sum(int i) {
        long long ret = 0;
        while (i > 0) {
            ret += node[i];
            i -= i & -i;
        }
        return ret;
    }
};

long double solvex() {
  long double l = -INF;
  long double r = INF;
  rep(_, 100) {
    BIT bit(n);
    long double m = (r + l) / 2.0;
    int cnt = 0;
    vector<pair<long double, int>> v;
    rep(i, n) {
      long double y = (-a[i] / b[i]) * m + c[i] / b[i];
      v.push_back({y, i});
    }
    sort(all(v));
    for (auto p : v) {
      int idx = posx[p.se];
      cnt += bit.sum(idx);
      bit.add(idx, 1);
    }
    //cout << m << " " << cnt << endl;
    //dump(v);
    if (n % 2 == 0) {
      if (cnt  < n * (n - 1) / 4) l = m;
      else r = m;
    }
    else {
      if (cnt < n * (n - 1) / 4 + 1) l = m;
      else r = m;
    }
  }
  return r;
}

long double solvey() {
  long double l = -INF;
  long double r = INF;
  rep(_, 100) {
    BIT bit(n);
    long double m = (r + l) / 2.0;
    int cnt = 0;
    vector<pair<long double, int>> v;
    rep(i, n) {
      long double x = (-b[i] / a[i]) * m + c[i] / a[i];
      v.push_back({x, i});
    }
    sort(all(v));
    for (auto p : v) {
      int idx = posy[p.se];
      cnt += bit.sum(idx);
      bit.add(idx, 1);
    }
    //cout << m << " " << cnt << endl;
    if ((n * (n - 1) / 2) % 2 == 0) {
      if (cnt  < n * (n - 1) / 4) l = m;
      else r = m;
    }
    else {
      if (cnt < n * (n - 1) / 4 + 1) l = m;
      else r = m;
    }
  }
  return r;
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n;
    rep(i, n) {
      cin >> a[i] >> b[i] >> c[i];
      fx.push_back({-a[i] / b[i], i});
      fy.push_back({-b[i] / a[i], i});
    }
    sort(all(fx));
    sort(all(fy));
    int i = 1;
    for (auto p : fx) {
      posx[p.se] = i;
      i++;
    }
    i = 1;
    for (auto p : fy) {
      posy[p.se] = i;
      i++;
    }
    cout << fixed << setprecision(20) << solvex() << " " << solvey() << endl;
}