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.StringTokenizer;
  
  public class test {
  
      public static void main(String[] args) {
          FastScanner sc = new FastScanner();
          int t = sc.ni();
          while (t-- > 0) {
              long n = sc.nl();
              long[][] p = new long[(int)n][2];
              long[] a = new long[(int)n + 1];
              long[] b = new long[(int)n + 1];
              for (int i = 0; i < n; i++) {
                  p[i][0] = sc.nl();
                  p[i][1] = sc.nl();
                  a[(int)p[i][0]]++;
                  b[(int)p[i][1]]++;
              }
              long sum = n * (n - 1) * (n - 2) / 6;
              long minus = 0;
              for (int i = 0; i < n; i++) {
                  minus += (a[(int)p[i][0]] - 1) * (b[(int)p[i][1]] - 1);
              }
              sum -= minus;
              System.out.println(sum);
          }
      }
  
      static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public FastScanner() {
              br = new BufferedReader(new InputStreamReader(System.in), 32768);
              st = null;
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int ni() {
              return Integer.parseInt(next());
          }
  
          int[][] graph(int N, int[][] edges) {
              int[][] graph = new int[N][];
              int[] sz = new int[N];
              for (int[] e: edges) {
                  sz[e[0]] += 1;
                  sz[e[1]] += 1;
              }
              for (int i = 0; i < N; i++) {
                  graph[i] = new int[sz[i]];
              }
              int[] cur = new int[N];
              for (int[] e: edges) {
                  graph[e[0]][cur[e[0]]] = e[1];
                  graph[e[1]][cur[e[1]]] = e[0];
                  cur[e[0]] += 1;
                  cur[e[1]] += 1;
              }
              return graph;
          }
  
          int[] intArray(int N, int mod) {
              int[] ret = new int[N];
              for (int i = 0; i < N; i++)
                  ret[i] = ni()+mod;
              return ret;
          }
  
          long nl() {
              return Long.parseLong(next());
          }
  
          long[] longArray(int N, long mod) {
              long[] ret = new long[N];
              for (int i = 0; i < N; i++)
                  ret[i] = nl()+mod;
              return ret;
          }
  
          double nd() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  