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 PyPy3 (2.4.0) (judge verdict: Accepted)

#設定
import sys
input = sys.stdin.buffer.readline

#ライブラリインポート
from collections import defaultdict

#入力受け取り
def getlist():
	return list(map(int, input().split()))

class UnionFind:
	def __init__(self, n):
		self.par = [i for i in range(n)]
		self.rank = [0] * n
		self.size = [1] * n

	def find(self, x):
		if self.par[x] == x:
			return x
		else:
			self.par[x] = self.find(self.par[x])
			return self.par[x]

	def same_check(self, x, y):
		return self.find(x) == self.find(y)

	def union(self, x, y):
		x = self.find(x)
		y = self.find(y)
		if self.rank[x] < self.rank[y]:
			if self.same_check(x, y) != True:
				self.size[y] += self.size[x]
				self.size[x] = 0
			self.par[x] = y
		else:
			if self.same_check(x, y) != True:
				self.size[x] += self.size[y]
				self.size[y] = 0
			self.par[y] = x
			if self.rank[x] == self.rank[y]:
				self.rank[x] += 1

	def siz(self, x):
		x = self.find(x)
		return self.size[x]

#処理内容
def main():
	N = int(input())
	Lx = []
	Ly = []
	UF = UnionFind(N)
	for i in range(N):
		x, y = getlist()
		Lx.append([x, y, i])
		Ly.append([y, x, i])
	Lx = sorted(Lx)
	Ly = sorted(Ly)
	for i in range(N - 1):
		if Lx[i][0] == Lx[i + 1][0]:
			UF.union(Lx[i][2] , Lx[i + 1][2])
		if Ly[i][0] == Ly[i + 1][0]:
			UF.union(Ly[i][2] , Ly[i + 1][2])
	for i in range(N):
		UF.par[i] = UF.find(i)

	nodeX = [[] for i in range(N)] 
	nodeY = [[] for i in range(N)]
	PAR = UF.par
	for i in range(N):
		nodeX[PAR[Lx[i][2]]].append(Lx[i][0])
		nodeY[PAR[Lx[i][2]]].append(Lx[i][1])
	# print(nodeX)
	# print(nodeY)

	#計算
	ans = 0
	D = defaultdict(int)
	for i in range(N):
		if D[PAR[i]] == 0:
			D[PAR[i]] = 1
			ans += len(set(nodeX[PAR[i]])) * len(set(nodeY[PAR[i]])) - UF.size[PAR[i]]
	print(ans) 



if __name__ == '__main__':
	main()