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 itertools import groupby
from operator import itemgetter

N = int(input())
x, y = (
    zip(*(map(int, input().split()) for _ in range(N))) if N else
    ((), ())
)


class UnionFindTree:
    def __init__(self, n):
        self.p = [i for i in range(n + 1)]
        self.r = [0 for _ in range(n + 1)]

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

    def union(self, x, y):
        px = self.find(x)
        py = self.find(y)
        if px != py:
            if self.r[px] < self.r[py]:
                self.p[px] = py
            else:
                self.p[py] = px
                if self.r[px] == self.r[py]:
                    self.r[px] += 1

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


# x座標またはy座標を共有する点についてグループ化する
# グループごとに可能な操作回数を求めるのは簡単
uft = UnionFindTree(N)
for d in range(2):
    for _, g in groupby(
                sorted(
                    zip(
                        x,
                        y,
                        range(1, N + 1)
                    ),
                    key=itemgetter(d)
                ),
                key=itemgetter(d)
            ):
        p = list(g)
        for (_, _, i), (_, _, j) in zip(p[1:], p):
            uft.union(i, j)
ans = sum(
    len({x for x, _, _ in C}) * len({y for _, y, _ in C}) - len(list(C))
    for C in (
        list(g)
        for _, g in groupby(
            sorted(
                (
                    (x, y, uft.find(i))
                    for x, y, i in zip(
                        x,
                        y,
                        range(1, N + 1)
                    )
                ),
                key=itemgetter(2)
            ),
            key=itemgetter(2)
        )
    )
)

print(ans)
