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
# reference -> https://atcoder.jp/contests/abc131/submissions/10358162
import sys
sys.setrecursionlimit(100000)

# 再帰的に連結部分の辺の数を求める
# 同じ辺を2回ずつ数えるので、最終的には2で割る
def solve(x, x_set, y_set, x2y, y2x):
  num_e = 0
  if x in x2y:
    x_set.add(x)
    yl = x2y.pop(x)
    num_e += len(yl)
    for y in yl:
      num_e += solve(y, y_set, x_set, y2x, x2y)
  return num_e


N = int(input())
X2Y = {}
Y2X = {}
for i in range(N):
  x, y = map(int, input().split())
  if x in X2Y:
    X2Y[x].append(y)
  else:
    X2Y[x] = [y]
  if y in Y2X:
    Y2X[y].append(x)
  else:
    Y2X[y] = [x]


ans = 0
while X2Y:
    x = next(iter(X2Y)) # まだスタートに選んでいないxを取り出す
    x_set = set() # 上のxと連結であるxの集合
    y_set = set() # 上のxと連結であるyの集合
    num_e = solve(x, x_set, y_set, X2Y, Y2X)
    ans += len(x_set)*len(y_set) - num_e//2

print(ans)