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: Accepted)

#include<bits/stdc++.h>
#define N 100005
#define LL long long
using namespace std;
template<class T>inline void Rd(T &x){
	x=0;char c=getchar();bool f=0;
	while(c<'0'||c>'9'){if(c=='-')f^=1;c=getchar();}
	while(c>='0'&&c<='9')x=(x<<1)+(x<<3)+c-'0',c=getchar();
	if(f)x=-x;
}
int n;
int X[N],Y[N];
int fa[N<<1];
int get(int x){return x==fa[x]?x:fa[x]=get(fa[x]);}
LL cnt[2][N<<1];

int main(){
	Rd(n);
	for(register int i=1;i<=(N<<1);++i)fa[i]=i;
	for(register int i=1,x,y;i<=n;++i){
		Rd(x),Rd(y);
		fa[get(x)]=get(y+N);
	}
	for(register int i=1;i<=N;++i)++cnt[0][get(i)];
	for(register int i=1+N;i<=(N<<1);++i)++cnt[1][get(i)];
	LL sum=0;
	for(register int i=1;i<=(N<<1);++i)
		sum+=cnt[0][i]*cnt[1][i];
	cout<<sum-n<<endl;
	return 0;
}