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 <vector>
#include <list>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <queue>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <fstream>
#include <cstdio>
#include <climits>
#include <complex>
#include <cstdint>
#include <tuple>

#define M_PI       3.14159265358979323846

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }
inline int readInt() { int x; scanf("%d", &x); return x; }

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;

//container util

//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())

//repetition
//------------------------------------------
#define FOR(i,s,n) for(int i=s;i<(int)n;++i)
#define REP(i,n) FOR(i,0,n)
#define MOD 1000000007

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
struct Edge {
    int to, cost;
    Edge(int to, int cost): to(to), cost(cost) {}
};

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vector<Edge>> AdjList;
AdjList graph;

const int INF = 100000000;

int N, M;

map<int, int> mp1, mp2;
vector<pair<int, int>> vp;
int main() {
    //cout << fixed << setprecision(15);

    int n, m, x, y; cin >> n;

    REP(i, n){
        cin >> x >> y;
        vp.push_back({x, y});
    }

   int minc = 10000000;
   for(int i=0; i<n; ++i){
       for(int j=0; j<n; ++j){
           if(i==j) continue;
            int dx = vp[j].first - vp[i].first;
            int dy = vp[j].second - vp[i].second;


            for(int k=0; k<n; ++k){
                int cost = 1;
                if(k==i || k==j) continue;
                for(int l=0; l<n; ++l){
                    if(l==k || l==i || l==j) continue;
                    int ndx = vp[l].first - vp[k].first;
                    int ndy = vp[l].second - vp[k].second;

                    if(dx == ndx && dy == ndy){

                    }else{
                        cost++;
                    }
                }
                if(vp[k].first - vp[j].first == dx && vp[k].second - vp[j].second == dy){

                }else{
                    cost++;
                }
                minc = min(minc, cost);
            }
       }
   }
   if(minc == 10000000) minc = 1;
    cout << minc << endl;

    return 0;
}