Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in Python (3.4.3) (judge verdict: Time Limit Exceeded)

N = int(input())

V = 100005
to = [[] for _ in range(V * 2)]
visited = [False for _ in range(V * 2)]
cnt = []

def dfs(v):  
  L = [v]
  while len(L) > 0:
    l = L.pop(0)
    if visited[l]:
      continue
    visited[l] = True
    cnt[l // V] += 1
    for u in to[l]:
      L.append(u)

for i in range(N):
  x, y = map(int, input().split())
  y += V
  to[x].append(y)
  to[y].append(x)

ans = 0
for i in range(V * 2):
  if visited[i]:
    continue
  cnt = [0, 0]
  dfs(i)
  ans += cnt[0] * cnt[1]
ans -= N
print(ans)