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>
using namespace std;
#define reg register
#define MaxN 100003
#define fghiudsh dfguyshyi
#define fers dfguyshyi
#define fersgv dfguyshyi
int N;
int fa[MaxN];
long long ans, cnty[MaxN], cntx[MaxN];
map<int, bool> bk[MaxN];
struct Nd {
	int x, y;
}a[MaxN];

inline int Rd() {
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }
	while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
	return x * f;
}
inline int Find(int d) {
	if(fa[d] == d) return d;
	else return fa[d] = Find(fa[d]);
} 
inline bool cmp(Nd aa, Nd bb) {
	return aa.x < bb.x;
}

int main() {
	N = Rd();
	for(reg int i = 1; i <= 100000; ++i)
		fa[i] = i;
	for(reg int i = 1; i <= N; ++i)
		a[i].x = Rd(), a[i].y = Rd();
	sort(a + 1, a + 1 + N, cmp);
	for(reg int i = 1; i < N; ++i)
		if(a[i].x == a[i + 1].x)
			fa[Find(a[i].y)] = Find(a[i + 1].y);
	for(reg int i = 1; i <= 100000; ++i)
		++cnty[Find(i)];
	for(reg int i = 1; i <= N; ++i)
		if(!bk[Find(a[i].y)][a[i].x]) {
			++cntx[Find(a[i].y)];
			bk[Find(a[i].y)][a[i].x] = true;
		}
	for(reg int i = 1; i <= 100000; ++i)
		ans += cnty[i] * cntx[i];
	cout << ans - N << endl;
	return 0;
}