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 (7.3.0) (judge verdict: Wrong Answer)

import sys
from operator import itemgetter
input = sys.stdin.readline

class UnionFind:
    def __init__(self, n):
        self.n = n
        self.par = [-1]*(n+1)#それぞれの要素がどの要素の子であるか
        self.cycle = [i for i in range(n+1)]

    def find(self, x):
        if self.par[x] < 0:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]#それぞれの要素の根を再帰的に求める

    def union(self, x, y):
        self.cycle[x], self.cycle[y] = self.cycle[y], self.cycle[x]
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        if self.par[x] > self.par[y]:
            x, y = y, x
        
        self.par[x] += self.par[y]
        self.par[y] = x

    def same(self, x, y):
        return self.find(x) == self.find(y)#x,yが同じ集合に属するかどうか
    
    def size(self, x):
        return -self.par[self.find(x)]
    
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if root == self.find(i)]        

def compress(lis, n=2, last_sort=False):
    length = len(lis)
    first_value = -1
    for i in range(n):
        lis.sort(key=itemgetter(i))
        value = first_value
        index = 0
        for j in range(length):
            if lis[j][i] != value:
                index += 1
            value = lis[j][i]
            lis[j][i] = index
    
    if last_sort: lis.sort()
    return lis
                

def main():
    n = int(input())
    p = [list(map(int, input().split())) for _ in range(n)]
    
    compress(p, last_sort=True)
    
    key = [set() for _ in range(n+1)]
    for i in range(n):
        key[p[i][1]].add(p[i][0])
    
    union = UnionFind(n)
    for i in range(n-1):
        if p[i][0] == p[i+1][0]:
            union.union(p[i][1], p[i+1][1])
    
    already = [False]*(n+1)
    
    ans = 0
    for i in range(1, n+1):
        if already[i]:
            continue
        index = i
        table = [index]
        while True:
            already[index] = True
            index = union.cycle[index]
            if index == i:
                break
            table.append(index)
        judge = set()
        field = 0
        for v in table:
            judge = judge | key[v]
            field -= len(key[v])
        field += len(table)*len(judge)
        ans += field
        
    print(ans)
            
        
    
    
if __name__ == "__main__":
    main()