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
# -*- coding: utf-8 -*-
import sys
from collections import defaultdict
sys.setrecursionlimit(10**9)
INF=10**18
MOD=10**9+7
input=lambda: sys.stdin.readline().rstrip()
YesNo=lambda b: bool([print('Yes')] if b else print('No'))
YESNO=lambda b: bool([print('YES')] if b else print('NO'))
int1=lambda x:int(x)-1

N=int(input())

class UnionFind:
    def __init__(self,n):
        self.par=[i for i in range(n)]
        self.siz=[1]*n
        
    def root(self,x):
        while self.par[x]!=x:
            self.par[x]=self.par[self.par[x]]
            x=self.par[x]
        return x
    
    def unite(self,x,y):
        x=self.root(x)
        y=self.root(y)
        if x==y:
            return False
        if self.siz[x]<self.siz[y]:
            x,y=y,x
        self.siz[x]+=self.siz[y]
        self.par[y]=x
        return True
    
    def is_same(self,x,y):
        return self.root(x)==self.root(y)
    
    def size(self,x):
        return self.siz[self.root(x)]
MAX=10**5
u=UnionFind(MAX*2)
for _ in range(N):
    x,y=map(int1,input().split())
    u.unite(x,y+MAX)
lx,ly=[0]*MAX*2,[0]*MAX*2
for i in range(MAX):
    lx[u.root(i)]+=1
    ly[u.root(i+MAX)]+=1
ans=0
for i in range(MAX*2):
    ans+=lx[i]*ly[i]
print(ans-N)