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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

using ll = long long int;

const int MAX = (int)(1e5 + 5);
const ll INF = (ll)(1e10 + 5);

const int MAX_N = (int)(1e5 + 5);

int n;
int x[MAX_N], y[MAX_N];

vector<int> graph[MAX * 2 + 20];

int node2conn[MAX * 2 + 20];

void dfs(int node, int conn) {
  if (node2conn[node] >= 0) return;  // visited

  node2conn[node] = conn;

  for (auto &child : graph[node]) {
    dfs(child, conn);
  }
}

int main(void) {
  // Here your code !

  scanf("%d", &n);

  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &(x[i]), &(y[i]));

    int node_x = x[i];
    int node_y = y[i] + MAX;

    graph[node_x].push_back(node_y);
    graph[node_y].push_back(node_x);
  }

  for (int i = 0; i < MAX * 2 + 20; ++i) {
    node2conn[i] = -1;
  }

  for (int i = 0; i < n; ++i) {
    int node_x = x[i];
    dfs(node_x, i);
  }

  unordered_set<int> xs;
  for (int i = 0; i < n; ++i) {
    int node_x = x[i];
    xs.insert(node_x);
  }

  unordered_map<int, ll> conn2xnodenum;
  for (auto &e : xs) {
    conn2xnodenum[node2conn[e]] += 1LL;
  }

  unordered_set<int> ys;
  for (int i = 0; i < n; ++i) {
    int node_y = y[i] + MAX;
    ys.insert(node_y);
  }

  unordered_map<int, ll> conn2ynodenum;
  for (auto &e : ys) {
    conn2ynodenum[node2conn[e]] += 1LL;
  }

  unordered_set<int> conns;
  for (int i = 0; i < MAX * 2 + 20; ++i) {
    int conn = node2conn[i];
    if (conn < 0) continue;
    conns.insert(conn);
  }

  ll ans = 0LL;

  for (auto &key : conns) {
    ans += conn2xnodenum[key] * conn2ynodenum[key];
  }

  ans -= (ll)(n);

  printf("%lld\n", ans);

  return 0;
}