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
n=int(input())
xy=[]
X=[]
Y=[]
for _ in range(n):
  x,y=map(int,input().split())
  xy.append([x,y])
  X.append(x)
  Y.append(y)
X=list(set(X))
Y=list(set(Y))
X.sort()
Y.sort()
from bisect import bisect_left
xy=[[bisect_left(X,x),bisect_left(Y,y)] for x,y in xy]
mx=0
my=0
for x,y in xy:
  mx=max(mx,x)
  my=max(my,y)
mx,my=mx+1,my+1
bg=[[] for _ in range(mx+my)]
for x,y in xy:
  y+=mx
  bg[x].append(y)
  bg[y].append(x)
mi=set(range(mx+my))
ren=[]
while mi:
  todo=[mi.pop()]
  rx,ry=0,0
  while todo:
    v=todo.pop()
    if v<mx:
      rx+=1
    else:
      ry+=1
    for nv in bg[v]:
      if nv in mi:
        todo.append(nv)
        mi.discard(nv)
  ren.append([rx,ry])
ans=0
for rx,ry in ren:
  ans+=rx*ry
ans-=n
print(ans)
