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
input = sys.stdin.readline

N = int(input())
XY = [[set() for i in range(10**5+1)] for j in range(2)]
for i in range(N):
  x, y = map(int, input().split())
  XY[0][x].add(y)
  XY[1][y].add(x)

is_visited_x = [False] * (10**5 + 1)
def dfs(x):
  global is_visited_x
  is_visited_x[x] = True
  stack = [[x], []]
  idx = 0
  num_e = 0
  num_xy = [1, 0]
  parent = 0
  while stack[0] or stack[1]:
    node = stack[idx][-1]
    children = XY[idx][node]
    if children:
      child = children.pop()
      XY[idx-1][child].remove(node) # 今来た道を削除
      num_e += 1
      if child in stack[idx-1]:
        continue
      stack[idx-1].append(child)
      num_xy[idx-1] += 1
      is_visited_x[child] += idx
      idx = (idx + 1) % 2
    else:
      stack[idx].pop()
      idx = (idx + 1) % 2
  return num_e, num_xy

ans = 0
for x in range(1, 10**5+1):
  if is_visited_x[x]:
    continue
  num_e, num_xy = dfs(x)
  ans += num_xy[0]*num_xy[1] - num_e

print(ans)