Problem:
Score : 600 points

Problem Statement

There are N dots in a two-dimensional plane. The coordinates of the i-th dot are (x_i, y_i).

We will repeat the following operation as long as possible:

Choose four integers a, b, c, d (a \neq c, b \neq d) such that there are dots at exactly three of the positions (a, b), (a, d), (c, b) and (c, d), and add a dot at the remaining position.

We can prove that we can only do this operation a finite number of times. Find the maximum number of times we can do the operation.

Constraints

1 \leq N \leq 10^5

1 \leq x_i, y_i \leq 10^5

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 maximum number of times we can do the operation.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define REP(i,n) for(long long i=0;i<n;i++)
#define REP1(i,n) for(long long i=1;i<=n;i++)
#define REP2D(i,j,h,w) for(long long i=0;i<h;i++) for(long long j=0;j<w;j++)
#define REP2D1(i,j,h,w) for(long long i=1;i<=h;i++) for(long long j=1;j<=w;j++)
#define PER(i,n) for(long long i=((n)-1);i>=0;i--)
#define PER1(i,n) for(long long i=(n);i>0;i--)
#define FOR(i,a,b) for(long long i=(a);i<(b);i++)
#define FORE(i,a,b) for(long long i=(a);i<=(b);i++)
#define ITE(arr) for(auto ite=(arr).begin();ite!=(arr).end();++ite)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define YNPRT(b) cout<<((b)?"Yes":"No")<<endl
#define REV(arr) reverse(ALL(arr))
#define PRT(a) cout<<a<<endl;
#define PRTLST(arr,num) REP(_i,num) cout<<_i<<" - "<<arr[_i]<<endl;
#define PRTLST2(arr2,d1,d2) REP(_i,d1) REP(_j,d2) cout<<_i<<","<<_j<<" : "<<arr2[_i][_j]<<endl;
#define PRTLST2D(arr2,d1,d2) do{cout<<"L\t";REP(_i,d2) cout<<_i<<"\t"; cout<<endl; REP(_i,d1){cout<<_i<<"\t";REP(_j,d2){cout<<arr2[_i][_j]<<"\t";}cout<<endl;}}while(0);
#define fst first
#define scd second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const int INF_INT = 2147483647;
const ll INF_LL = 9223372036854775807LL;
const ull INF_ULL = 18446744073709551615Ull;
const ll P = 92540646808111039LL;
const int Move[4][2] = {-1,0,1,0,0,1,0,-1};
const int Move_[8][2] = {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};

//---------------------
#define MAXN 100000
//---------------------

template<typename T>
ostream& operator<<(ostream& os,const vector<T>& v) {ITE(v) os << *ite << " ";return os;}


ll n,x[MAXN],y[MAXN];
vector<pll> points;
int main(){
	cin>>n;
	REP(i,n){
		cin>>x[i]>>y[i];
		points.push_back(pll(x[i],y[i]));
	}
	if(points.size()<=2) {cout<<0<<endl;return 0;}
	sort(ALL(points));
	ll id=0,lastx=-1;
	set<ll> beforey;
	vector<ll> samex;
	ll ret=0;
	while(1){
	  lastx=points[id].fst;
	  while(points[id].fst==lastx)
	  {
	  	samex.push_back(points[id].scd);
	  	id++;
	  	if(id>=n) break;
	  }
	  
	  if(!beforey.empty()){
	    ll sameynum=0;
	  	ITE(samex){
	  		if(beforey.count(*ite)){
	  		  	sameynum++;
	  			break;
	  		}
	  	}
	  	ret+=samex.size()-2*sameynum+beforey.size();
	  }
	  
	  REP(i,samex.size()){
	  	beforey.insert(samex[i]);
	  }
	  samex.clear();
	  if(id>=n) break;
	}
	
	cout<<ret<<endl;
	
	return 0;
}