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 sys
  from sys import stdin
  
  tt = int(stdin.readline())
  
  ANS = []
  for loop in range(tt):
  
      n = int(stdin.readline())
  
      ab = []
      anum = {}
      bnum = {}
  
      for i in range(n):
          a,b = map(int,stdin.readline().split())
          ab.append((a,b))
  
          if a not in anum:
              anum[a] = 0
          if b not in bnum:
              bnum[b] = 0
  
          anum[a] += 1
          bnum[b] += 1
  
  
      twonl = 0
      for a in anum:
          c = anum[a]
          twonl += c * (c-1) // 2 * (n-c)
      for b in bnum:
          c = bnum[b]
          twonl += c * (c-1) // 2 * (n-c)
  
      tans = 0
      for a,b in ab:
          tans += (anum[a]-1) * (bnum[b]-1)
  
      aans = 0
      for a in anum:
          c = anum[a]
          aans += c * (c-1) * (c-2) // 6
      for b in bnum:
          c = bnum[b]
          aans += c * (c-1) * (c-2) // 6
  
      #print (twonl , tans , ans)
      aans = n*(n-1)*(n-2)//6  - (twonl - tans + aans) #both
  
  
      #a only
      dpA = [0,0,0]
      for a in anum:
          c = anum[a]
          dpA[2] += dpA[1] * c
          dpA[1] += dpA[0] * c
          dpA[0] += c
  
      dpB = [0,0,0]
      for b in bnum:
          c = bnum[b]
          dpB[2] += dpB[1] * c
          dpB[1] += dpB[0] * c
          dpB[0] += c
  
      twoans = dpA[-1] + dpB[-1]
      ANS.append(str(twoans - aans))
  
  print ("\n".join(ANS))