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: #!/usr/bin/env python3
  # from typing import *
  
  import sys
  import io
  import math
  import collections
  import decimal
  import itertools
  import bisect
  import heapq
  
  
  def input():
      return sys.stdin.readline()[:-1]
  
  
  # sys.setrecursionlimit(1000000)
  
  # _INPUT = """2
  # 4
  # 2 4
  # 3 4
  # 2 1
  # 1 3
  # 5
  # 1 5
  # 2 4
  # 3 3
  # 4 2
  # 5 1
  
  # """
  # sys.stdin = io.StringIO(_INPUT)
  
  INF = 10**10
  
  def solve(N, T):
      counter1 = collections.Counter([a for a, b in T])
      counter2 = collections.Counter([b for a, b in T])
  
      v1 = 0
      for a, n in counter1.items():
          if n >= 2:
              v1 += n*(n-1)//2 * (N-n)
  
      v2 = 0
      for a, n in counter1.items():
          if n >= 3:
              v2 += n*(n-1)*(n-2)//6
  
      v3 = 0
      for a, b in T:
          v3 += (counter1[a]-1) * (counter2[b]-1)
      
      v4 = 0
      for a, n in counter2.items():
          if n >= 2:
              v4 += n*(n-1)//2 * (N-n)
  
      v5 = 0
      for a, n in counter2.items():
          if n >= 3:
              v5 += n*(n-1)*(n-2)//6
  
      m = N*(N-1)*(N-2)//6
      ans = (m-v1-v2) + (m-v4-v5) - (m-v1-v2+v3-v4-v5)
      return ans
  
  
  T0 = int(input())
  for _ in range(T0):
      N = int(input())
      T = []
      for _ in range(N):
          a, b = map(int, input().split())
          T.append((a, b))
      print(solve(N, T))
  