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 deque
N = int(input())
V = set()
E = {}

for _ in range(N):
  x,y = map(int, input().split())
  y += 100010
  V.add(x)
  V.add(y)
  if x not in E:
    E[x] = [y]
  else:
    E[x].append(y)
  if y not in E:
    E[y] = [x]
  else:
    E[y].append(x)

#print(V)
#print(E)
## DFS
visited = [ False ] * 200020
willSearch = [ False ] * 200020
Q = deque()
ans = 0
for v in V:
  if visited[v]:
    continue

  Q.append(v)
  xcount, ycount = 0,0
  edge_count = 0
  while len(Q) > 0:
    now = Q.pop()
    visited[now] = True

    if now > 100005:
      ycount += 1
    else:
      xcount += 1
      
    for nxt in E[now]:
      edge_count += 1
      if visited[nxt] or willSearch[nxt]:
        continue
      #print(v,now,nxt)

      willSearch[nxt] = True

      Q.append(nxt)
  #print(v,xcount, ycount, edge_count)
  ans += xcount * ycount
  ans -= edge_count // 2
  
print(ans)