#include <bits/stdc++.h>
using namespace std;

const double EPS = 1e-8;
const double INF = 1e12;
typedef complex<double> P;
namespace std {
  bool operator < (const P& a, const P& b) {
    return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
  }
}
double cross(const P& a, const P& b) {
  return imag(conj(a)*b);
}
double dot(const P& a, const P& b) {
  return real(conj(a)*b);
}

struct L : public vector<P> {
  L(const P &a, const P &b) {
    push_back(a); push_back(b);
  }
};

int ccw(P a, P b, P c) {
  b -= a; c -= a;
  if (cross(b, c) > 0)   return +1;       // counter clockwise
  if (cross(b, c) < 0)   return -1;       // clockwise
  if (dot(b, c) < 0)     return +2;       // c--a--b on line
  if (norm(b) < norm(c)) return -2;       // a--b--c on line
  return 0;
}
typedef vector<P> G;

struct C {
  P p; double r;
  C(const P &p, double r) : p(p), r(r) { }
};

bool intersectLL(const L &l, const L &m) {
  return abs(cross(l[1]-l[0], m[1]-m[0])) > EPS || // non-parallel
         abs(cross(l[1]-l[0], m[0]-l[0])) < EPS;   // same line
}

bool intersectLS(const L &l, const L &s) {
  return cross(l[1]-l[0], s[0]-l[0])*       // s[0] is left of l
         cross(l[1]-l[0], s[1]-l[0]) < EPS; // s[1] is right of l
}
bool intersectSS(const L &s, const L &t) {
  return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 &&
         ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}

int main(void){
    int n;
    cin >> n;
    vector<L> lines;
    int a,b,c,d;
    for(int i=0;i<n;i++){
        cin >> a >>  b >> c >> d;
        P p1(a,b),p2(c,d);
        L l(p1,p2);
        lines.push_back(l);
    }

    vector<pair<int,int>> lp;
    for(int i=-101;i<=101;i++){
        lp.push_back(make_pair(i,-101));
        lp.push_back(make_pair(i,101));
        lp.push_back(make_pair(-101,i));
        lp.push_back(make_pair(101,i));
    }
    int ret =0;
    for(int i=0;i<lp.size();i++){
        for(int j=0;j<lp.size();j++){
        	if(i == j)continue;
            int cnt =0;
            P p1(lp[i].first,lp[i].second),p2(lp[j].first,lp[j].second);
            L l(p1,p2);
            for(int k=0;k<n;k++){
                if(intersectSS(l,lines[k])){
                    cnt++;
                }
            }
            ret = max(ret,cnt);
        }
    }

    cout << ret << endl;
    return 0;
}