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 sun.awt.windows.WBufferStrategy;
  
  import java.io.*;
  import java.util.*;
  
  
  public class MainSolve {
      public static void main(String[] args) throws IOException {
          BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));
          BufferedWriter out=new BufferedWriter(new OutputStreamWriter(System.out));
          int testNum=Integer.parseInt(reader.readLine());
          for (int i = 0; i < testNum; i++) {
              int n=Integer.parseInt(reader.readLine());
              int[] countA=new int[n+1];
              int[] countB=new int[n+1];
              int[][] arr=new int[n][2];
              for (int j = 0; j < n; j++) {
                  String aAndB=reader.readLine();
                  String []AB=aAndB.split(" ");
                  arr[j][0]=Integer.parseInt(AB[0]);
                  arr[j][1]=Integer.parseInt(AB[1]);
                  countA[Integer.parseInt(AB[0])]++;
                  countB[Integer.parseInt(AB[1])]++;
              }
              out.write(solve(countA,countB,n,arr)+"\n");
              ans.clear();
              indexArray.clear();
          }
          out.flush();
          out.close();
          reader.close();
      }
  
      public static String solve(int[]a,int[] b,int n,int[][] arr){
          long sum=n*1L*(n-1)*(n-2)/6;
          for (int i = 0; i < n; i++) {
              sum-=(long) (a[arr[i][0]]-1)*(b[arr[i][1]]-1);
          }
          return String.valueOf(sum);
      }
  
      private static List<List<int []>> ans=new ArrayList<>();
      private static List<int []> indexArray=new ArrayList<>();
  
      private static void dfs(int[][] a,int index){
          if (indexArray.size()==3){
              boolean zero=indexArray.get(0)[0]!=indexArray.get(1)[0]&&indexArray.get(1)[0]!=indexArray.get(2)[0]&&indexArray.get(0)[0]!=indexArray.get(2)[0];
              boolean one=indexArray.get(0)[1]!=indexArray.get(1)[1]&&indexArray.get(1)[1]!=indexArray.get(2)[1]&&indexArray.get(0)[1]!=indexArray.get(2)[1];
              if ((zero||one)&&!ans.contains(new ArrayList<>(indexArray) )){
                  ans.add(new ArrayList<>(indexArray));
                  return;
              }
          }
          for (int i = index; i < a.length; i++) {
              indexArray.add(a[i]);
              dfs(a,i+1);
              indexArray.remove(indexArray.size()-1);
          }
      }
  
  }
  
  