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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.List;
  import java.util.StringTokenizer;
  
  /**
   * Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
   * <p>
   * To modify the template, go to Preferences -> Editor -> File and Code Templates -> Other
   */
  
  public class Main {
      static class Solution {
          private int n;
          private int[] C;
          private int[] N;
          private long[][] ch;
          private long[] P;
          private int[] match;
          public Solution(int n, List<List<Integer>> points) {
              this.n = n;
              C = new int[n];
              N = new int[n];
              for (List<Integer> pr: points) {
                  int i = pr.get(0), j = pr.get(1);
                  C[j]++;
                  N[i]++;
              }
              precomputeP();
              precomputeChoose();
              precomputeMatch(points);
          }
          private void precomputeP() {
              P = new long[n];
              int tot = 0;
              for (int i = n-1; i>=0; i--) {
                  if (i==n-1) {
                      P[i] = 0;
                  } else  {
                      P[i] = P[i+1];
                  }
                  P[i] += (long)(tot)*(long)N[i];
                  tot += N[i];
              }
          }
          private long choose(int n, int k) {
              if (k>n) return 0;
              return ch[n][k];
          }
          private void precomputeChoose() {
              ch = new long[n+1][4];
  
              ch[0][0] = 1;
              for (int num = 1; num<=n; num++) {
                  ch[num][0] = 1;
                  for (int k = 1; k<=3; k++) {
                      ch[num][k] = ch[num-1][k-1] + ch[num-1][k];
                  }
              }
          }
          private void precomputeMatch(List<List<Integer>> points) {
              match = new int[n];
              for (List<Integer> pr: points) {
                  int i = pr.get(0), j = pr.get(1);
  //                assert(C[j]>0);
                  match[i] += C[j]-1;
              }
          }
          public long solve() {
              long ans = 0;
  
              for (int i = 0; i<n; i++) {
                  long Tiii = choose(N[i],3);
                  long Ti = 0;
                  if (i+1<n) Ti = N[i] * P[i+1];
  
  //                assert (n-N[i]-match[i] >= 0);
                  long Tii = (long)(n-N[i]-match[i])*choose(N[i],2) + (long)(match[i])*choose(N[i]-1,2);
                  ans += Tiii+Tii+Ti;
              }
              return ans;
          }
          /* The hard case is
           * 2 of the same topic "i"
           * Note if 2 are of the same topic, we MUST
           * satisfy the 3 unique difficulties constraint
           * We can NOT do Choose(N[i],2) * (n - N[i])
           * Because any random (n-N[i]) choose for the third
           * might have the same difficulty as one of the 2 from row i
           *
           *
           * */
  
  
      }
  
      public static void main(String[] args) {
          // Write your solution here
          FastScanner fs = new FastScanner();
          int T = fs.nextInt();
          while (T > 0) {
              int n = fs.nextInt();
              final List<List<Integer>> points = new ArrayList<>();
              for (int t=0; t<n; t++) {
                  int i = fs.nextInt(), j = fs.nextInt();
                  i--; j--;
                  points.add(Arrays.asList(i,j));
              }
              System.out.println((new Solution(n,points)).solve());
              T--;
          }
      }
  
  
      static class FastScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
  }