Problem:
Score : 300 points

Problem Statement

There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i).

We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation:

Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1.

Find the minimum total cost required to collect all the balls when we optimally choose p and q.

Constraints

1 \leq N \leq 50

|x_i|, |y_i| \leq 10^9

If i \neq j, x_i \neq x_j or y_i \neq y_j.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the minimum total cost required to collect all the balls.

Sample Input 1

2
1 1
2 2

Sample Output 1

1

If we choose p = 1, q = 1, we can collect all the balls at a cost of 1 by collecting them in the order (1, 1), (2, 2).

Sample Input 2

3
1 4
4 6
7 8

Sample Output 2

1

If we choose p = -3, q = -2, we can collect all the balls at a cost of 1 by collecting them in the order (7, 8), (4, 6), (1, 4).

Sample Input 3

4
1 1
1 2
2 1
2 2

Sample Output 3

2

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

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using vb=vector<bool>;
using vvb=vector<vb>;
using vd=vector<double>;
using vvd=vector<vd>;
using vi=vector<int>;
using vvi=vector<vi>;
using vl=vector<ll>;
using vvl=vector<vl>;
using pii=pair<int,int>;
using pll=pair<ll,ll>;
using tll=tuple<ll,ll>;
using tlll=tuple<ll,ll,ll>;
using vs=vector<string>;

#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rep(i,n) range(i,0,n)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define range(i,a,n) for(int i=(a);i<(n);i++)

#define LINF ((ll)1ll<<60)
#define INF ((int)1<<30)
#define EPS (1e-9)
#define MOD (1000000007ll)
#define fcout(a) cout<<setprecision(a)<<fixed
#define fs first
#define sc second
#define PI 3.1415926535897932384

int dx[]={1,0,-1,0,1,-1,-1,1},dy[]={0,1,0,-1,1,1,-1,-1};

template<class T>bool chmax(T&a,T b){if(a<b){a=b; return true;}return false;}
template<class T>bool chmin(T&a,T b){if(a>b){a=b; return true;}return false;}
template<class S>S acm(vector<S>&a){return accumulate(all(a),S());}
template<class S>S max(vector<S>&a){return *max_element(all(a));}
template<class S>S min(vector<S>&a){return *min_element(all(a));}
void YN(bool b){cout<<(b?"YES":"NO")<<"\n";}
void Yn(bool b){cout<<(b?"Yes":"No")<<"\n";}
void yn(bool b){cout<<(b?"yes":"no")<<"\n";}

ll max(int a,ll b){return max((ll)a,b);} ll max(ll a,int b){return max(a,(ll)b);}
template<class T>void puta(T&&t){cout<<t<<"\n";}
template<class H,class...T>void puta(H&&h,T&&...t){cout<<h<<' ';puta(t...);}
template<class S,class T>ostream&operator<<(ostream&os,pair<S,T>p){os<<"["<<p.first<<", "<<p.second<<"]";return os;};
template<class S>auto&operator<<(ostream&os,vector<S>t){bool a=1; for(auto s:t){os<<(a?"":" ")<<s;a=0;} return os;}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n,ans=INF;
	cin>>n;
	vl x(n),y(n);
	rep(i,n)cin>>x[i]>>y[i];
	rep(i,n){
		rep(j,i){
			set<ll> s;
			ll p=x[i]-x[j];
			ll q=y[i]-y[j];
			rep(k,n){
				if(p==0)s.insert(x[k]);
				else s.insert(p*y[k]-x[k]*q);
			}
			chmin(ans,(int)s.size());
		}
	}
	if(n==1)ans=1;
	cout<<ans<<endl;
}