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
from queue import PriorityQueue
import sys

sys.setrecursionlimit(10**9)

IA = lambda: [int(x) for x in input().split()]
IM = lambda N: [IA() for _ in range(N)]
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
N=100005
fa=[int(0) for i in range(2*(N+1))]
xNum=[int(0) for i in range(2*(N+1))]
yNum=[int(0) for i in range(2*(N+1))]

def find(x):
    if(x==fa[x]):return x
    else:
        fa[x]=find(fa[x])
        return fa[x]

n=input()
n=int(n)

for i in range(2*N+1):
    fa[i]=i

for i in range(n):
    x,y=IA()
    y+=N
    fx=int(find(x))
    fy=int(find(y))
 
    if(fx!=fy):
        fa[fy]=fx
for i in range(N):
    xNum[find(i)]+=1
for i in range(N):
    yNum[find(i+N)]+=1
ans=0
for i in range(N):
    ans+=xNum[i]*yNum[i]
print(ans-n)


