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>
#define rep(i,l,r) for (ll i=l;i<=r;i++)
using namespace std;
typedef long long ll;
struct st1{ll x,y;};
ll a,b,c,d,ans,N,K,fax[100007],fay[100007];
ll nx[100007],ny[100007];
st1 q[100007];
ll get(ll *fa,ll x1){
	if (fa[x1]==x1) return x1;
	fa[x1]=get(fa,fa[x1]);
	return fa[x1];
}
bool cmp1(st1 x1,st1 x2){return x1.x<x2.x;}
bool cmp2(st1 x1,st1 x2){return x1.y<x2.y;}
bool cmp3(st1 x1,st1 x2){
	return x1.x<x2.x||(x1.x==x2.x&&x1.y<x2.y);
}
int main(){
	scanf("%lld",&N);
	rep(i,1,100000) fax[i]=i,fay[i]=i,nx[i]=1,ny[i]=1;
	rep(i,1,N) scanf("%lld%lld",&q[i].x,&q[i].y);
	sort(q+1,q+N+1,cmp1);
	rep(i,1,N) if (q[i].x==q[i-1].x){
		a=get(fay,q[i-1].y);b=get(fay,q[i].y);
		ny[b]+=ny[a];fay[a]=b;
	}
	sort(q+1,q+N+1,cmp2);
	rep(i,1,N) if (q[i].y==q[i-1].y){
		a=get(fax,q[i-1].x);b=get(fax,q[i].x);
		nx[b]+=nx[a];fax[a]=b;
	}
	rep(i,1,N) q[i].x=get(fax,q[i].x),q[i].y=get(fay,q[i].y);
	rep(i,1,N) if (q[i].x!=q[i-1].x||q[i].y!=q[i-1].y) ans+=nx[q[i].x]*ny[q[i].y];
	
	printf("%lld\n",ans-N);
	return 0;
}