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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<iostream>
#include<vector>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<string.h>
#include<functional>
#include<limits.h>
#include<stdlib.h>
#include<cmath>
#include<cstring>
#include<set>
#include<climits>
#include<deque>
#include <iomanip>
#include<utility>
using namespace std;

#define intmax INT_MAX
#define lmax LONG_MAX
#define uintmax UINT_MAX
#define ulmax ULONG_MAX
#define llmax LLONG_MAX
#define ll long long
#define rep(i,a,N) for((i)=(a);(i)<(N);(i)++)
#define rrp(i,N,a) for((i)=(N)-1;(i)>=(a);(i)--)
#define llfor ll i,j
#define sc(a) cin>>a
#define pr(a) cout<<a<<endl
#define pY puts("YES")
#define pN puts("NO")
#define py puts("Yes")
#define pn puts("No")
#define pnn printf("\n")
#define all(a) a.begin(),a.end()
#define push(a,b) (a).push_back(b)
#define llvec vector<vector<ll>>
#define charvec vector<vector<char>>
#define size(a,b) (a,vector<ll>(b))
#define llpvec vector<pair<ll,ll>>
#define S 21 
llfor;ll mod=1000000007;
/*pi*/double pi=acos(-1);
/*繰り上げ除算*/ll cei(ll x,ll y){ll ans=x/y;if(x%y!=0)ans++;return ans;}
/*最大公約数*/ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
/*最小公倍数*/ll lcm(ll x,ll y){return x/gcd(x,y)*y;}
/*n乗*/ll llpow(ll x,ll n){ll ans=1;rep(i,0,n){ans*=x;ans%=mod;}return ans;}
/*階乗*/ll fact(ll x){ll ans=1;rep(i,0,x)ans*=(x-i);return ans;}
/*nCr*/ll ncr(ll n,ll r){return fact(n)*fact(n-r)/fact(r);}
/*nPr*/ll npr(ll n,ll r){return fact(n)/fact(n-r);}
/*primejudge*/bool prime(ll a){if(a<=1)return false;for(i=2;i*i<=a;i++){if(a%i==0)return false;}return true;}
/*Fibonacci数列*/ll fib(ll x){ll fibo[x+10];fibo[0]=1;fibo[1]=1;fibo[2]=1;rep(i,3,x+1)fibo[i]=fibo[i-1]+fibo[i-2];return fibo[x];}
/*桁数*/ll dig(ll n){ll d=1,tmp=n;while(tmp/10>0){tmp/=10;d++;}return d;}
/*行列のコピー*/void matcpy(ll a[S][S],ll b[S][S]){ll t=i;rep(i,0,S)rep(j,0,S)b[i][j]=a[i][j];/*(aをbに)*/i=t;}
/*行列のリセット*/void matrst(ll a[S][S]){ll t=i;rep(i,0,S)rep(j,0,S)a[i][j]=0;i=t;}
/*行列の2乗*/void matpow(ll a[S][S]){ll ans[S][S];matrst(ans);ll t=i,k;rep(i,0,S)rep(j,0,S)rep(k,0,S)ans[i][j]+=a[i][k]*a[k][j];matcpy(ans,a);i=t;}
/*行列の乗算*/void matmul(ll a[S][S],ll b[S][S], ll as[S][S]){ll t=i,k;matrst(as);rep(i,0,S)rep(j,0,S)rep(k,0,S)as[i][j]+=a[i][k]*b[k][j];i=t;}
//ll ans=0;
////////////////////////////////////////////////////////////
int main(){
 ll n;
 cin>>n;
 ll a[n+10],b[n+10]; 
 rep(i,0,n)cin>>a[i]>>b[i];
 map<pair<ll,ll>,ll>m;
 ll p,q;
 ll c=0;
 rep(i,0,n){
  rep(j,0,n){
   if(i==j)continue;
   p=a[i]-a[j];
   q=b[i]-b[j];
   m[{p,q}]++;
   c=max(c,m[{p,q}]);
  }
 }
 ll ans=n-c;
 if(n==1)ans=1;
 pr(ans);
return 0;}