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
class UF():
  def __init__(self, N):
    self._parent=[n for n in range(0, N)]
    self._size=[1] * N

  def find_root(self, x):
    if self._parent[x]==x:return x
    self._parent[x]=self.find_root(self._parent[x])
    return self._parent[x]

  def unite(self, x, y):
    gx=self.find_root(x)
    gy=self.find_root(y)
    if gx==gy:return

    if self._size[gx]<self._size[gy]:
      self._parent[gx]=gy
      self._size[gy]+=self._size[gx]
    else:
      self._parent[gy]=gx
      self._size[gx]+=self._size[gy]

  def size(self, x):
    return self._size[self.find_root(x)]

  def samegroup(self, x, y):
    return self.find_root(x)==self.find_root(y)

  def groupnum(self):
    N=len(self._parent)
    ans=0
    for i in range(N):
      if self.find_root(i)==i:
        ans+=1
    return ans

import sys
input=sys.stdin.readline
n=int(input())
UF=UF(2*10**5)
for i in range(n):
  x,y=map(int,input().split())
  UF.unite(x-1+10**5,y-1)
UF.groupnum()
ct=[[0,0] for i in range(2*10**5)]
for i in range(2*10**5):
  if i<10**5:
    ct[UF._parent[i]][0]+=1
  else:
    ct[UF._parent[i]][1]+=1
ans=0
for i in range(2*10**5):
  ans+=ct[i][0]*ct[i][1]
print(ans-n)