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
U = 10 ** 5
root = [None] * (2*U+1)
N = int(input())
XY = [[int(x) for x in input().split()] for _ in range(N)]

def find_root(x):
  y = root[x]
  if y is None:
    root[x] = x
    return x
  if x == y:
    return y
  z = find_root(y)
  root[x] = z
  return z

def merge(x,y):
  rx = find_root(x)
  ry = find_root(y)
  if rx == ry:
    return
  root[ry] = rx
  return

for x,y in XY:
  merge(x,U+y)

pts_cnt = defaultdict(int)
x_cnt = defaultdict(int)
y_cnt = defaultdict(int)

for x in range(U+1):
  if root[x] is None:
    continue
  x_cnt[find_root(x)] += 1

for y in range(U+1,U+U+1):
  if root[y] is None:
    continue
  y_cnt[find_root(y)] += 1

for x,y in XY:
  pts_cnt[find_root(x)] += 1

answer = 0
for k,v in pts_cnt.items():
  answer += x_cnt[k] * y_cnt[k] - v

print(answer)