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: Accepted)


#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <fstream>
#include <sstream>
#include <set>
#include <map>
#include <vector>
#include <list>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <bitset>     //UWAGA - w czasie kompilacji musi byc znany rozmiar wektora - nie mozna go zmienic
#include <cassert>
#include <iomanip>        //do setprecision
#include <ctime>
#include <complex>
using namespace std;

#define FOR(i,b,e) for(int i=(b);i<(e);++i)
#define FORQ(i,b,e) for(int i=(b);i<=(e);++i)
#define FORD(i,b,e) for(int i=(b)-1;i>=(e);--i)
#define REP(x, n) for(int x = 0; x < (n); ++x)

#define ST first
#define ND second
#define PB push_back
#define PF push_front
#define MP make_pair
#define LL long long
#define ULL unsigned LL
#define LD long double
#define pii pair<int,int>
#define pll pair<LL,LL>

const double pi = 3.14159265358979323846264;
const int mod=1000000007;

int main(){
	 //cin.tie(0);
	//ios::sync_with_stdio(false);
	//std::cout << std::fixed;
	//std::cout << std::setprecision(12);
	//std::cout << std::defaultfloat;

	int n;
	cin>>n;
	vector<LL> x(n),y(n);
	FOR(i,0,n)cin>>x[i]>>y[i];
	vector<pll> xy(n);
	FOR(i,0,n){
		xy[i]=MP(x[i],y[i]);
	}
	sort(xy.begin(),xy.end());
	LL ans=0;
	FOR(i,0,n){
		FOR(j,i+1,n){
			LL p=xy[j].ST-xy[i].ST,q=xy[j].ND-xy[i].ND;
			LL cnt=0;
			vector<int> used(n,0);
			FOR(k,0,n){
				LL nx=xy[k].ST,ny=xy[k].ND;
				if(used[k])continue;
				used[k]=1;
				FOR(kk,k+1,n){
					LL tx=xy[kk].ST,ty=xy[kk].ND;
					if(tx-nx==p&&ty-ny==q){
						used[kk]=1;
						cnt++;
						nx=tx,ny=ty;
					}
				}
			}
			//cerr<<p<<" "<<q<<" "<<cnt<<endl;
			ans=max(ans,cnt);
		}
	}
	cout<<n-ans<<endl;

	return 0;
}