/*
【题目】POJ 1118
【题意】判断多个点中最多有多少个点共线
【题解】按横纵坐标大小排序,从小到大遍历每一个点,
计算该点与剩余点的斜率,记录到数组中
在数组中查找有有多少个斜率相同的点
注意只有两个点的情况,应该输出2
*/
#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
const int inf=0x3f3f3f;
struct Point{
double x,y;
Point(double x=0,double y=0):x(x),y(y){}
};
Point a[1000];
typedef Point Vector;
Vector operator + (Vector A,Vector B){
return Vector(A.x+B.x,A.y+B.y);
}
Vector operator - (Vector A,Vector B){
return Vector(A.x-B.x,A.y-B.y);
}
Vector operator * (Vector A,double p){
return Vector(A.x*p,A.y*p);
}
Vector operator / (Vector A,double p){
return Vector(A.x/p,A.y/p);
}
bool operator < (const Point &a,const Point &b){
if(a.x==b.x){
return a.y<b.y;
}
return a.x<b.x;
}
const double eps=1e-6;
int sgn(double x){
if(fabs(x)<eps){
return 0;
}
if(x<0){
return -1;
}
return 1;
}
bool operator == (const Point &a,const Point &b){
if(sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0){
return true;
}
return false;
}
double pos[1000];
int main(){
while(1){
int n;
cin>>n;
if(n==0){
break;
}
for(int i=0;i<n;i++){
cin>>a[i].x>>a[i].y;
}
sort(a,a+n);
int ans=1;
for(int i=0;i<n;i++){
int cur=0;
for(int j=i+1;j<n;j++){
if(a[j].x-a[i].x==0){
pos[cur++]=inf;
}
else{
pos[cur++]=(a[j].y-a[i].y)/(a[j].x-a[i].x);
}
}
sort(pos,pos+cur);
int cnt=1;
for(int j=1;j<cur;j++){
if(fabs(pos[j]-pos[j-1])<1e-6){
cnt++;
ans=max(ans,cnt);
}else{
cnt=1;
}
}
}
cout<<ans+1<<endl;
}
return 0;
}