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
  input = sys.stdin.buffer.readline 
  
  def process(A):
      n = len(A)
      total = n*(n-1)*(n-2)//6
      pairs = {}
      topic = {}
      diff = {}
      for i in range(n):
          a, b = A[i]
          if (a, b) not in pairs:
              pairs[(a, b)] = 0
          pairs[(a, b)]+=1
          if a not in topic:
              topic[a] = 0
          topic[a]+=1
          if b not in diff:
              diff[b] = 0
          diff[b]+=1
      bad = 0
      for x in pairs:
          a, b = x
          y = pairs[x]
          z1 = topic[a]
          z2 = diff[b]
          #one topic, one difficulty
          bad_triple1 = y*(y-1)*(y-2)//6
        #  print(bad_triple, 'One')
          bad+=bad_triple1
          
          #one topic, two difficulties
          bad_triple2 = (y*(y-1)//2)*(z1-2)-y*(y-1)*(y-2)//6
        #  print(bad_triple, 'Two')
          bad+=bad_triple2
          
          #two topics, one difficulty
          bad_triple3 = (y*(y-1)//2)*(z2-2)-y*(y-1)*(y-2)//6
        #  print(bad_triple, "Three")
          bad+=bad_triple3
      #two topics, two difficulties
      #case 1
      #a x
      #a x
      #b y
          unmatched_pairs = (n-z1-z2+y)
          bad_triple4 = ((y*(y-1))//2)*unmatched_pairs
     #     print(bad_triple, 'Four')
          bad+=bad_triple4
          #two topics, two difficulties
          #case 2
          #a x
          #a y
          #b x  
          other_a = (z1-y)
          other_b = (z2-y)
          bad_triple5 = y*other_a*other_b
       #   print(bad_triple, "Five")
          bad+=bad_triple5
      return total-bad
  
  t = int(input())
  for i in range(t):
      n = int(input())
      A = []
      for j in range(n):
          a, b = [int(x) for x in input().split()]
          A.append([a, b])
      answer = process(A)
      sys.stdout.write(str(answer)+'\n')