Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 1
5 1
5 5

Sample Output 1

1

By choosing a = 1, b = 1, c = 5, d = 5, we can add a dot at (1, 5). We cannot do the operation any more, so the maximum number of operations is 1.

Sample Input 2

2
10 10
20 20

Sample Output 2

0

There are only two dots, so we cannot do the operation at all.

Sample Input 3

9
1 1
2 1
3 1
4 1
5 1
1 2
1 3
1 4
1 5

Sample Output 3

16

We can do the operation for all choices of the form a = 1, b = 1, c = i, d = j (2 \leq i,j \leq 5), and no more. Thus, the maximum number of operations is 16.

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

import collections

class UnionFind:

    def __init__(self, size):
        # 負の値はルート (集合の代表) で集合の個数
        # 正の値は次の要素を表す
        self.size = size
        self.table = [-1 for _ in range(size)]

    def find(self, x):
        # 集合の代表を求める
        cs = []
        while self.table[x] >= 0:
            cs.append(x)
            x = self.table[x]
        for c in cs:
            self.table[c] = x
        return x

    def union(self, x, y):
        # 併合
        s1 = self.find(x)
        s2 = self.find(y)
        if s1 != s2:
            if self.table[s1] >= self.table[s2]:
                self.table[s1] += self.table[s2]
                self.table[s2] = s1
            else:
                self.table[s2] += self.table[s1]
                self.table[s1] = s2
        return self.table[s1]



n = int(input())
ys = set()
y2xs = collections.defaultdict(set)

for i in range(n):
    x,y = map(int, input().split())
    ys.add(y)
    y2xs[y].add(x)
ys = sorted(list(ys))
y_cnt = [len(y2xs[y]) for y in ys]
y_size = len(ys)
y_used = [False] * y_size

total = 0


uf = UnionFind(y_size)
for i in range(y_size-1):
    for j in range(i+1, y_size):
        if y2xs[ys[uf.find(i)]] & y2xs[ys[uf.find(j)]]:
            if not y_used[i]:
                total -= len(y2xs[ys[i]])
                y_used[i] = True
            if not y_used[j]:
                total -= len(y2xs[ys[j]])
                y_used[j] = True
            
            new_set = y2xs[ys[uf.find(i)]] | y2xs[ys[uf.find(j)]]
            uf.union(i, j)
            y2xs[ys[uf.find(i)]] = new_set

for i in range(y_size):
    if y_used[uf.find(i)]:
        total += len(y2xs[ys[uf.find(i)]])

print(total)