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.readline
  def iinput(): return int(input())
  def sinput(): return input().rstrip()
  def i0input(): return int(input()) - 1
  def linput(): return list(input().split())
  def liinput(): return list(map(int, input().split()))
  def miinput(): return map(int, input().split())
  def li0input(): return list(map(lambda x: int(x) - 1, input().split()))
  def mi0input(): return map(lambda x: int(x) - 1, input().split())
  INF = 10**20
  MOD = 1000000007
  
  # Hopcroft-Karp Algorithm
  # from collections import deque
  # class HopcroftKarp:
  #     def __init__(self, N0, N1):
  #         self.N0 = N0
  #         self.N1 = N1
  #         self.N = N = 2+N0+N1
  #         self.G = [[] for i in range(N)]
  #         for i in range(N0):
  #             forward = [2+i, 1, None]
  #             forward[2] = backward = [0, 0, forward]
  #             self.G[0].append(forward)
  #             self.G[2+i].append(backward)
  #         self.backwards = bs = []
  #         for i in range(N1):
  #             forward = [1, 1, None]
  #             forward[2] = backward = [2+N0+i, 0, forward]
  #             bs.append(backward)
  #             self.G[2+N0+i].append(forward)
  #             self.G[1].append(backward)
  
  #     def add_edge(self, fr, to):
  #         #assert 0 <= fr < self.N0
  #         #assert 0 <= to < self.N1
  #         v0 = 2 + fr
  #         v1 = 2 + self.N0 + to
  #         forward = [v1, 1, None]
  #         forward[2] = backward = [v0, 0, forward]
  #         self.G[v0].append(forward)
  #         self.G[v1].append(backward)
  
  #     def bfs(self):
  #         G = self.G
  #         level = [None]*self.N
  #         deq = deque([0])
  #         level[0] = 0
  #         while deq:
  #             v = deq.popleft()
  #             lv = level[v] + 1
  #             for w, cap, _ in G[v]:
  #                 if cap and level[w] is None:
  #                     level[w] = lv
  #                     deq.append(w)
  #         self.level = level
  #         return level[1] is not None
  
  #     def dfs(self, v, t):
  #         if v == t:
  #             return 1
  #         level = self.level
  #         for e in self.it[v]:
  #             w, cap, rev = e
  #             if cap and level[v] < level[w] and self.dfs(w, t):
  #                 e[1] = 0
  #                 rev[1] = 1
  #                 return 1
  #         return 0
  
  #     def flow(self):
  #         flow = 0
  #         G = self.G
  #         bfs = self.bfs; dfs = self.dfs
  #         while bfs():
  #             *self.it, = map(iter, G)
  #             while dfs(0, 1):
  #                 flow += 1
  #         return flow
  
  #     def matching(self):
  #         return [cap for _, cap, _ in self.backwards]
  
  t = iinput()
  from collections import defaultdict
  for _ in [0] * t:
      n = iinput()
      A = defaultdict(int)
      B = defaultdict(int)
      all = []
      # g = HopcroftKarp(n, n)
      for _ in [0] * n:
          a, b = li0input()
          A[a] += 1
          B[b] += 1
          all.append((a, b))
      #     g.add_edge(a, b)
      # joint = g.flow()
      # joint = joint * (joint - 1) * (joint - 2) // 6
      ans1 = n * (n - 1) * (n - 2) // 6
      ans2 = n * (n - 1) * (n - 2) // 6
      for v in A.values():
          if v >= 2:
              ans1 -= (n - v) * v * (v - 1) // 2
          if v >= 3:
              ans1 -= v * (v - 1) * (v - 2) // 6
      for v in B.values():
          if v >= 2:
              ans2 -= (n - v) * v * (v - 1) // 2
          if v >= 3:
              ans2 -= v * (v - 1) * (v - 2) // 6
      bar_joint = 0
      for a, b in all:
          bar_joint += max(0, A[a] - 1) * max(0, B[b] - 1)
      joint = ans1 + ans2 + bar_joint - n * (n - 1) * (n - 2) // 6
      print(ans1 + ans2 - joint)