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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

/**
 * Created at 22:27 on 2019-06-22
 */

import java.awt.*;
import java.io.*;
import java.util.*;

public class Main {

  static FastScanner sc = new FastScanner();

  static int[] dx = {0, 1, 0, -1};
  static int[] dy = {-1, 0, 1, 0};

  static long MOD = (long) (1e9 + 7);

  public static void main(String[] args) {

    int N = sc.nextInt();

    Dot[] dots = new Dot[N];

    for (int i=0; i<N; i++) {
      dots[i] = new Dot(i, sc.nextInt(), sc.nextInt());
    }

    UnionFindTree tree = new UnionFindTree(N);

    Arrays.sort(dots, (o1, o2) -> o1.x - o2.x);

    for (int i=1; i<N; i++) {
      if (dots[i-1].x == dots[i].x) {
        tree.union(dots[i-1].n, dots[i].n);
      }
    }

    Arrays.sort(dots, (o1, o2) -> o1.y - o2.y);

    for (int i=1; i<N; i++) {
      if (dots[i-1].y == dots[i].y) {
        tree.union(dots[i-1].n, dots[i].n);
      }
    }

    HashSet<Integer>[] tmpX = new HashSet[N];
    HashSet<Integer>[] tmpY = new HashSet[N];

    for (int i=0; i<N; i++) {
      tmpX[i] = new HashSet();
      tmpY[i] = new HashSet();
    }

    for (int i=0; i<N; i++) {
      int index = tree.find(i);
      tmpX[index].add(dots[i].x);
      tmpY[index].add(dots[i].y);
    }

    int ans = -N;
    for (int i=0; i<N; i++) {
      ans += tmpX[i].size()*tmpY[i].size();
    }

    System.out.println(ans);

  }

  public static class Dot {
    int n, x, y;
    public Dot(int n, int x, int y) {
      this.n = n;
      this.x = x;
      this.y = y;
    }
  }

  public static class UnionFindTree {

    int[] parent; //インデックスにとノードを対応させ、そのルートノードのインデックスを格納
    int[] rank; //parentと同様に、木の高さを格納

    public UnionFindTree(int size) {
      this.parent = new int[size];
      this.rank = new int[size];

      for (int i = 0; i < size; i++) {
        makeSet(i);
      }
    }

    public void makeSet(int i) {
      parent[i] = i;
      rank[i] = 0; //集合の高さ
    }

    public void union(int x, int y) {
      int xRoot = find(x);
      int yRoot = find(y);

      //xが属する木の方が大きい場合
      if (rank[xRoot] > rank[yRoot]) {
        parent[yRoot] = xRoot; //yの親をxに更新

      } else if(rank[xRoot] < rank[yRoot]) {
        parent[xRoot] = yRoot;

      } else if (xRoot != yRoot){
        parent[yRoot] = xRoot;
        rank[xRoot]++; //同じ高さの木がルートの子として着くから大きさ++;
      }
    }

    //引数の属する木のルートのidを返す
    public int find(int i) {
      if (i != parent[i]) {
        parent[i] = find(parent[i]);
      }
      return parent[i];
    }

    //同じ木に属しているか
    public boolean same(int x, int y) {
      return find(x) == find(y);
    }
  }

  public static class Mathf {

    public static int max(int[] a) {
      int M = a[0];
      for (int i = 1; i < a.length; i++) {
        M = Math.max(M, a[i]);
      }
      return M;
    }

    public static int min(int[] a) {
      int m = a[0];
      for (int i = 1; i < a.length; i++) {
        m = Math.min(m, a[i]);
      }
      return m;
    }

    public static long max(long[] a) {
      long M = a[0];
      for (int i = 1; i < a.length; i++) {
        M = Math.max(M, a[i]);
      }
      return M;
    }

    public static long min(long[] a) {
      long m = a[0];
      for (int i = 1; i < a.length; i++) {
        m = Math.min(m, a[i]);
      }
      return m;
    }

  }

  /*
    add()でインデックスを指定しない場合指定されたソート順に違わない位置に追加する
    (ただしソートされていることが前提となる)
    Comparatorが0を返したとき、それらの順序は保証しない
    (TreeSet, TreeMapと違い削除はしない)
   */
  static class TreeList<E> extends ArrayList<E> {

    Comparator<? super E> comparator;

    TreeList(Comparator<? super E> c) {
      super();
      comparator = c;
    }

    /*
      ソート済みのリストに要素を追加する
     */
    public boolean add(E e) {
      int lowIndex = 0;
      int highIndex = size() - 1;
      int index = 0;

      if (size() == 0) {
        super.add(e);
        return true;
      }

      if (comparator.compare(e, get(0)) < 0) {
        index = 0;
      } else if (comparator.compare(e, get(highIndex)) > 0) {
        index = highIndex + 1;
      } else {
        while (lowIndex <= highIndex) {

          if (highIndex == lowIndex + 1 || highIndex == lowIndex) {
            index = highIndex;
            break;
          }

          int midIndex = (lowIndex + highIndex) / 2;
          ;

          if (comparator.compare(e, get(midIndex)) > 0) {
            lowIndex = midIndex;
          } else {
            highIndex = midIndex;
          }

        }
      }

      super.add(index, e);
      return true;
    }

  }

  static class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
      if (ptr < buflen) {
        return true;
      } else {
        ptr = 0;
        try {
          buflen = in.read(buffer);
        } catch (IOException e) {
          e.printStackTrace();
        }
        if (buflen <= 0) {
          return false;
        }
      }
      return true;
    }

    private int readByte() {
      if (hasNextByte()) return buffer[ptr++];
      else return -1;
    }

    private static boolean isPrintableChar(int c) {
      return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
      while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
    }

    public boolean hasNext() {
      skipUnprintable();
      return hasNextByte();
    }

    public String next() {
      if (!hasNext()) throw new NoSuchElementException();
      StringBuilder sb = new StringBuilder();
      int b = readByte();
      while (isPrintableChar(b)) {
        sb.appendCodePoint(b);
        b = readByte();
      }
      return sb.toString();
    }

    public long nextLong() {
      if (!hasNext()) throw new NoSuchElementException();
      long n = 0;
      boolean minus = false;
      int b = readByte();
      if (b == '-') {
        minus = true;
        b = readByte();
      }
      if (b < '0' || '9' < b) {
        throw new NumberFormatException();
      }
      while (true) {
        if ('0' <= b && b <= '9') {
          n *= 10;
          n += b - '0';
        } else if (b == -1 || !isPrintableChar(b)) {
          return minus ? -n : n;
        } else {
          throw new NumberFormatException();
        }
        b = readByte();
      }
    }

    public int nextInt() {
      return (int) nextLong();
    }

    public int[] nextIntArray(int N) {
      int[] array = new int[N];
      for (int i = 0; i < N; i++) {
        array[i] = sc.nextInt();
      }
      return array;
    }

    public long[] nextLongArray(int N) {
      long[] array = new long[N];
      for (int i = 0; i < N; i++) {
        array[i] = sc.nextLong();
      }
      return array;
    }
  }

}