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: Accepted)

#include <bits/stdc++.h>
#define reg register
#define ll long long
#define int long long
#define ull unsigned long long
#define db double
#define pi pair<int, int>
#define pl pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vpi vector<pi>
#define vpl vector<pl>
#define pb push_back
#define er erase
#define SZ(x) (int) x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mkp make_pair
#define dmin(x, y) ((x) < (y) ? (x) : (y))
#define dmax(x, y) ((x) > (y) ? (x) : (y))
#define checkmax(x, y) ((x) < (y) ? ((x) = (y)) : (x))
#define checkmin(x, y) ((x) > (y) ? ((x) = (y)) : (x))
#define ms(data_name) memset(data_name, 0, sizeof(data_name))
#define msn(data_name, num) memset(data_name, num, sizeof(data_name))
#define For(i, j) for(reg int (i) = 1; (i) <= (j); ++(i))
#define For0(i, j) for(reg int (i) = 0; (i) < (j); ++(i))
#define Forx(i, j, k) for(reg int (i) = (j); (i) <= (k); ++(i))
#define Forstep(i , j, k, st) for(reg int (i) = (j); (i) <= (k); (i) += (st))
#define fOR(i, j) for(reg int (i) = (j); (i) >= 1; (i)--)
#define fOR0(i, j) for(reg int (i) = (j) - 1; (i) >= 0; (i)--)
#define fORx(i, j, k) for(reg int (i) = (k); (i) >= (j); (i)--)
#define tour(i, u) for(reg int (i) = head[(u)]; (i) != -1; (i) = nxt[(i)])
using namespace std;
char ch, B[1 << 20], *S = B, *T = B;
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 20, stdin), S == T) ? 0 : *S++)
#define isd(c) (c >= '0' && c <= '9')
int rdint() {
  int aa, bb;
  while(ch = getc(), !isd(ch) && ch != '-');
  ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
  while(ch = getc(), isd(ch))
    aa = aa * 10 + ch - '0';
  return bb ? aa : -aa;
}
ll rdll() {
  ll aa, bb;
  while(ch = getc(), !isd(ch) && ch != '-');
  ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
  while(ch = getc(), isd(ch))
    aa = aa * 10 + ch - '0';
  return bb ? aa : -aa;
}
const int MAXN = 1e5 + 10;
int n, f[(MAXN << 1) + 10], cnt1[(MAXN << 1) + 10], cnt2[(MAXN << 1) + 10];
int find(int x) {
  return f[x] == x ? x : f[x] = find(f[x]);
}
inline void work() {
  n = rdint();
  Forx(i, 0, MAXN << 1) {
    f[i] = i;
  }
  For(i, n) {
    int x = rdint(), y = rdint();
    f[find(x)] = f[find(y + MAXN)];
  }
  For(i, MAXN) {
    ++cnt1[find(i)];
  }
  Forx(i, MAXN + 1, MAXN << 1) {
    ++cnt2[find(i)];
  }
  ll ans = 0;
  For(i, MAXN << 1) {
    ans += cnt1[i] * cnt2[i];
  }
  printf("%lld\n", ans - n);
}
signed main() {
  // freopen("input.txt", "r", stdin);
  work();
  return 0;
}