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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define whole(x) (x).begin(),(x).end()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
using P = pair<int, int>;
#define debug(var) cerr << "[" << #var << "] " << var << endl
#define chmin(x, y) x = min(x, y)
const ll mod = 1000000007;
const int dx[] = {-1,0,1,0};
const int dy[] = {0,-1,0,1};
#define int ll

struct UnionFind {
    vector<int> d;
    UnionFind(int n=0): d(n, -1) {}
    int find(int x) {
        if (d[x]<0) return x;
        return d[x] = find(d[x]);
    }
    bool unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x==y) return false;
        if (d[x] > d[y]) swap(x, y);
        d[x] += d[y];
        d[y] = x;
        return true;
    }
    bool same(int x, int y) {
        return find(x) == find(y);
    }
    int size(int x) {
        return -d[find(x)];
    }
};
const int MX = 100005;
ll numedge[200005];


signed main(){
    int n;
    cin >> n;
    vector<P> v(n);
    rep(i, n) cin >> v[i].first >> v[i].second;
    UnionFind u(2*MX);
    rep(i, n) {
        int r1 = u.find(v[i].first);
        int r2 = u.find(v[i].second+MX);
        u.unite(v[i].first, v[i].second+MX);
        if (r1!=r2) {
            numedge[u.find(v[i].first)] = numedge[r1] + numedge[r2] + 1;
        }
        else {
            numedge[u.find(v[i].first)]++;
        }
    }
    vector<P> num(2*MX);
    
    rep(i, 2*MX) {
        int r = u.find(i);
        if (i/MX==0) {
            num[r].first++;
        }
        else {
            num[r].second++;
        }
    }

    ll ans = 0;

    rep(i, 2*MX) {
        //if (num[i].first>0) debug(num[i].first);
        //if (num[i].second>0) debug(num[i].second);
        //if (numedge[i]>0) debug(numedge[i]);
        if (u.size(i)) {
            ans += (ll)(num[i].first*num[i].second) - numedge[i];
        }
    }
    
    cout << ans << endl;
    return 0;
}