#pragma GCC optimize "O3"
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;

#define FOR(i,b,e) for(int i=b; i<=e; i++)
#define FORD(i,b,e) for(int i=b; i>=e; i--)
#define SIZE(x) ((int)x.size())
#define pb push_back
#define st first
#define nd second
#define sp ' '
#define ent '\n'

#define x st
#define y nd

int n;
set<ii> pts;
vii ha;

ii licz(ii a, ii b){
	return {a.x-b.x, a.y-b.y};
}

ii dodaj(ii a, ii b){
	return {a.x+b.x, a.y+b.y};
}

bool dziala(ii a){
	for(auto &x2: pts){
		if(pts.find(dodaj(x2, a)) == pts.end() && pts.find(licz(x2, a)) == pts.end()){
			return 0;
		}
	}
	return 1;
}

void solve(){
	int a, b, ans=0;
	cin>>n;
	FOR(i, 1, n){
		cin>>a>>b;
		pts.insert({a, b});
		ha.pb({a, b});
	}
	FOR(i, 2, n){
		if(ha[i-1]<ha[0]) swap(ha[i-1], ha[0]);
	}
	FOR(i, 2, n){
		ii pom=licz(ha[0], ha[i-1]);
		if(dziala(pom)) ans++;
	}
	cout<<ans*2<<ent;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	solve();
	return 0;
}

/*
5
0 0
1 1
2 2
2 0
3 1
*/