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>
// #include<boost/multiprecision/cpp_long.hpp>
using namespace std;
// using namespace boost::multiprecision;
#define BEGIN ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define END return EXIT_SUCCESS
#define rep(I,N) for(auto I=0;I<(N);++I)
#define up(I,A,B) for(auto I=(A);I<=(B);++I)
#define dw(I,A,B) for(auto I=(A);I>=(B);--I)
#define all(C) (C).begin(),(C).end()
#define rall(C) (C).rbegin(),(C).rend()
#define ft first
#define sd second
#define mp make_pair
#define mt make_tuple
#define pf push_front
#define pb push_back
#define pt pop_front
#define pk pop_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define rs resize
// #define long cpp_long
template<class T>inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;}
template<class T>inline void                                      in(T &p){cin>>p;}
template<class T1,class T2>inline void                            in(T1 &p1,T2 &p2){cin>>p1>>p2;}
template<class T1,class T2,class T3>inline void                   in(T1 &p1,T2 &p2,T3 &p3){cin>>p1>>p2>>p3;}
template<class T1,class T2,class T3,class T4>inline void          in(T1 &p1,T2 &p2,T3 &p3,T4 &p4){cin>>p1>>p2>>p3>>p4;}
template<class T1,class T2,class T3,class T4,class T5>inline void in(T1 &p1,T2 &p2,T3 &p3,T4 &p4,T5 &p5){cin>>p1>>p2>>p3>>p4>>p5;}
template<class T>inline void                                      ins(T &p){for_each(all(p),in<decltype(p[0])>);}
template<class T>inline void                             out(T p){cout<<p<<endl;}
template<class T1,class T2>inline void                   out(T1 p1,T2 p2){cout<<p1<<" "<<p2<<endl;}
template<class T1,class T2,class T3>inline void          out(T1 p1,T2 p2,T3 p3){cout<<p1<<" "<<p2<<" "<<p3<<endl;}
template<class T1,class T2,class T3,class T4>inline void out(T1 p1,T2 p2,T3 p3,T4 p4){cout<<p1<<" "<<p2<<" "<<p3<<" "<<p4<<endl;}
template<class T>inline                             void outs(T p){for_each(all(p),out<decltype(p[0])>);}

inline void solve(void){
  long N;
  in(N);
  vector<pair<long,long>>a,b;
  long x,y;
  rep(i,N){
    in(x,y);
    a.pb(mp(x,y));
    b.pb(mp(y,x));
  }
  sort(all(a));
  sort(all(b));
  map<pair<long,long>,long> c,d;
  rep(i,N-1){
    c[mp(a[i+1].ft-a[i].ft,a[i+1].sd-a[i].sd)]++;
    d[mp(b[i+1].ft-b[i].ft,b[i+1].sd-b[i].sd)]++;
  }
  long cnt=0;
  for(auto&& e:c)chmax(cnt,e.sd);
  for(auto&& e:d)chmax(cnt,e.sd);
  out(cnt!=1?N-cnt:N);
}
int main(int argc,char**argv){
  BEGIN;
  solve();
  END;
}