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
  
  IS_INTERACTIVE = False
  input = input  # type: ignore
  
  # input: function
  if not IS_INTERACTIVE:
      *data, = sys.stdin.read().split("\n")[::-1]
  
      def input():  # type: ignore
          return data.pop()
  
  def fprint(*args, **kwargs):
      print(*args, **kwargs, flush=True)
  
  def eprint(*args, **kwargs):
      print(*args, **kwargs, file=sys.stderr)
  
  '''
  1 1
  2 1
  2 2
  2 4
  3 4
  
  1-2-5
  1-3-5
  1-4-5
  
  
  
  '''
  
  
  
  def naive(topics, diffs):
      n = len(topics)
      cnt = 0
      for i in range(n):
          for j in range(i + 1, n):
              for k in range(j + 1, n):
                  if len({topics[i], topics[j], topics[k]}) == 3 or len({diffs[i], diffs[j], diffs[k]}) == 3:
                      cnt += 1
      return cnt
  
  
  def fast(topics, diffs):
      tcnt = [0] * (n + 1)
      dcnt = [0] * (n + 1)
      for t, d in zip(topics, diffs):
          tcnt[t] += 1
          dcnt[d] += 1
      pairs = list(zip(topics, diffs))
  
      ans = n * (n - 1) * (n - 2) // 6
      for i in range(n):
          c0 = tcnt[topics[i]]
          c1 = dcnt[diffs[i]]
          c0 -= 1
          c1 -= 1
          if c0 and c1:
              ans -= c0 * c1
  
      return ans
  
  
  '''
      n! / (n - 3)! / (3!)
  
      5 * 4 * 3
  
      1 2
      1 3
      2 2
  '''
  
  for _ in range(int(input())):
      n = int(input())
      topics = []
      diffs = []
      for _ in range(n):
          t, d = map(int, input().split())
          topics.append(t)
          diffs.append(d)
  
      print(fast(topics, diffs))
  
  
  
  