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
class UnionFind:

  def __init__(self, n):
    self.par = [i for i in range(n)]
    self.rank = [0] * n
    self.sizes = [1] * n

  def find(self, x):
    if self.par[x] != x:
      self.par[x] = self.find(self.par[x])
    return self.par[x]

  def union(self, x, y):
    x, y = self.find(x), self.find(y)
    if x == y:
      return False
    elif self.rank[x] < self.rank[y]:
      self.par[x] = y
      self.sizes[y] += self.sizes[x]
    elif self.rank[x] > self.rank[y]:
      self.par[y] = x
      self.sizes[x] += self.sizes[y]
    else:
      self.par[y] = x
      self.rank[x] += 1
      self.sizes[x] += self.sizes[y]
    return True

  def same(self, x, y):
    return self.find(x) == self.find(y)

  def size(self, x):
    return self.sizes[self.find(x)]

n = int(input())
nx = [0 for i in range(200020)]
ny = [0 for i in range(200020)]
uf = UnionFind(200020)
for i in range(n):
  x, y = list(map(int, input().split()))
  uf.union(x, y+100010)
ans = 0
for i in range(100010):
  nx[uf.find(i)] += 1
  ny[uf.find(i+100010)] += 1
for i in range(200020):
  ans += nx[i]*ny[i]
print(ans-n)