description: Monocarp is the coach of the Berland State University programming teams. He decided to compose a problemset for a training session for his teams.
  
  Monocarp has n problems that none of his students have seen yet. The i-th problem has a topic a_i (an integer from 1 to n) and a difficulty b_i (an integer from 1 to n). All problems are different, that is, there are no two tasks that have the same topic and difficulty at the same time.
  
  Monocarp decided to select exactly 3 problems from n problems for the problemset. The problems should satisfy at least one of two conditions (possibly, both):
  
    * the topics of all three selected problems are different; 
    * the difficulties of all three selected problems are different. 
  
  
  
  Your task is to determine the number of ways to select three problems for the problemset.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 50000) — the number of testcases.
  
  The first line of each testcase contains an integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of problems that Monocarp have.
  
  In the i-th of the following n lines, there are two integers a_i and b_i (1 ≤ a_i, b_i ≤ n) — the topic and the difficulty of the i-th problem.
  
  It is guaranteed that there are no two problems that have the same topic and difficulty at the same time.
  
  The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  Print the number of ways to select three training problems that meet either of the requirements described in the statement.
  
  Example
  
  Input
  
  
  2
  4
  2 4
  3 4
  2 1
  1 3
  5
  1 5
  2 4
  3 3
  4 2
  5 1
  
  
  Output
  
  
  3
  10
  
  Note
  
  In the first example, you can take the following sets of three problems:
  
    * problems 1, 2, 4; 
    * problems 1, 3, 4; 
    * problems 2, 3, 4. 
  
  
  
  Thus, the number of ways is equal to three.
solution: import java.util.*;
  import java.io.*;
  import java.math.*;
  import static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  
  public class Solution {
  
      public static long find(int n, long[] a, long[] b) {
          HashMap<Long, Long> m1 = new HashMap<>();
          HashMap<Long, Long> m2 = new HashMap<>();
          for (long l: a) {
              m1.put(l, m1.getOrDefault(l, 0L) + 1L);
          }
          for (long l: b) {
              m2.put(l, m2.getOrDefault(l, 0L) + 1L);
          }
  //        System.out.println(m1);
  //        System.out.println(m2);
          long res = ((long) n * (n - 1) * (n - 2)) / 6;
          for (int i = 0; i < a.length; i++) {
              if (m1.get(a[i]) > 1L && m2.get(b[i]) > 1L) {
  //                System.out.println("i:" + i);
  //                System.out.println(m1.get(a[i]));
  //                System.out.println(m2.get(b[i]));
                  res -= (m1.get(a[i]) - 1L) * (m2.get(b[i]) - 1L);
              }
          }
          return res;
      }
  
  
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter pw = new PrintWriter(System.out);
          int t = Integer.parseInt(br.readLine());
          for (int j = 0; j < t; j++) {
              int k = Integer.parseInt(br.readLine());
              long[] a = new long[k];
              long[] b = new long[k];
              for (int i = 0; i < k; i++) {
                  String[] s = br.readLine().split("\\s");
                  a[i] = Long.parseLong(s[0]);
                  b[i] = Long.parseLong(s[1]);
              }
              pw.println(find(k, a, b));
          }
          pw.flush();
      }
  }