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 collections import defaultdict
import sys
input = sys.stdin.readline
n = int(input())
xy = [list(map(int,input().split())) for i in range(n)]
graph = [[] for i in range(2*10**5+1)]
ans = 0
for x,y in xy:
  y += 10**5
  graph[x].append(y)
  graph[y].append(x)
  ans -= 1
vis = [0 for i in range(2*10**5+1)]
for i in range(1,2*10**5+1):
  if vis[i]:
    continue
  stack = [i]
  nx = 0
  ny = 0
  while stack:
    x = stack.pop()
    if x > 10**5:
      ny += 1
    else:
      nx += 1
    vis[x] = 1
    for y in graph[x]:
      if not vis[y]:
        stack.append(y)
        vis[y] = 1
  ans += nx*ny
print(ans)