#include<bits/stdc++.h>
#define Gc() getchar() 
#define Me(x,y) memset(x,y,sizeof(x))
#define Mc(x,y) memcpy(x,y,sizeof(x))
#define d(x,y) ((m)*(x-1)+(y))
#define R(n) (rnd()%(n)+1)
#define Pc(x) putchar(x)
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
using namespace std;using ll=long long;using db=double;using lb=long db;using ui=unsigned;using ull=unsigned long long;using pii=pair<int,int>;using LL=__int128;
const int N=2e5+5,M=N*100+5,K=600+5,mod=1e9+7,Mod=mod-1;const db eps=1e-9;const int INF=1e9+7;mt19937 rnd(263082);
int n;
struct vec{
	int x,y;
	vec operator -(const vec &B)const{return (vec){x-B.x,y-B.y};}
	ll operator *(const vec &B)const{return 1ll*x*B.y-1ll*y*B.x;}
}A[N];
void Solve(){
	int i,j;if(scanf("%d",&n)==-1) exit(0);
	for(i=1;i<=n;i++) scanf("%d%d",&A[i].x,&A[i].y);
	int ans=1;
	if(n==1) {printf("%d\n",1);return;}
	int T=50;while(T--){
		int x=R(n),y=R(n);while(x==y) y=R(n);
		int Ct=0;
		for(j=1;j<=n;j++) if((A[j]-A[x])*(A[y]-A[x])==0) Ct++;
		ans=max(ans,Ct);
	}
	if(ans<(n-ans)*2) printf("%d\n",n%3);
	else printf("%d\n",ans-2*(n-ans));
}
int main(){
	int t=1;
	// scanf("%d",&t);
	while(1) Solve();
	cerr<<clock()*1.0/CLOCKS_PER_SEC<<'\n';
}