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
import sys
sys.setrecursionlimit(1000000)

N = int(input().strip())

X = {}
Y = {}
G = (X, Y)

for _ in range(N):
  x, y = map(int, input().strip().split())
  if x not in X: X[x] = []
  if y not in Y: Y[y] = []
  X[x].append(y)
  Y[y].append(x)


def dfs_visit(visited, u, i):
  if u in visited[i]: return
  visited[i].add(u)
  for v in G[i][u]:
    dfs_visit(visited, v, (i + 1) % 2)


count = 0
i = 0
while len(X) > 0:
  visited = (set(), set())
  x = X.__iter__().__next__()
  dfs_visit(visited, x, 0)

  for x in visited[0]:
    count += len(visited[1]) - len(X.pop(x))

print(count)
