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
import bisect
import copy
import heapq
import math
import sys
from collections import *
from functools import lru_cache
from itertools import accumulate, combinations, permutations, product
def input():
    return sys.stdin.readline()[:-1]
def ruiseki(lst):
    return [0]+list(accumulate(lst))
sys.setrecursionlimit(5000000)
mod=pow(10,9)+7
al=[chr(ord('a') + i) for i in range(26)]
direction=[[1,0],[0,1],[-1,0],[0,-1]]

n=int(input())
xy=[list(map(int,input().split())) for i in range(n)]

dic={}
for i in range(n):
    x,y=xy[i]
    if (0,x) in dic:
        dic[(0,x)].append(y)
    else:
        dic[(0,x)]=[y]
    x,y=y,x
    if (1,x) in dic:
        dic[(1,x)].append(y)
    else:
        dic[(1,x)]=[y]
# print(dic)

lst=[[0]*(10**5+1) for i in range(2)]

ans=0
for x,y in xy:
    if lst[0][x]==1:
        continue
    lst[0][x]=1
    d=deque([(0,x)])
    xcnt,ycnt=0,0
    cnt=0
    while d:
        # print(d)
        now=d.popleft()
        if now[0]==0:
            xcnt+=1
        else:
            ycnt+=1
        for j in dic[now]:
            cnt+=1
            if lst[1-now[0]][j]==0:
                lst[1-now[0]][j]=1
                d.append((1-now[0],j))
    # print(xcnt,ycnt,cnt)
    ans+=xcnt*ycnt-cnt//2
print(ans)