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.

Examples

Input

3
1 1
5 1
5 5


Output

1


Input

2
10 10
20 20


Output

0


Input

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


Output

16
#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()