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 os,sys;from io import BytesIO, IOBase
  BUFSIZE = 8192
  class FastIO(IOBase):
      newlines = 0
      def __init__(self, file):
          self._fd = file.fileno();self.buffer = BytesIO();self.writable = "x" in file.mode or "r" not in file.mode;self.write = self.buffer.write if self.writable else None
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:break
              ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
      def readline(self):
          while self.newlines == 0:b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE));self.newlines = b.count(b"\n") + (not b);ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
      def flush(self):
          if self.writable:os.write(self._fd, self.buffer.getvalue());self.buffer.truncate(0), self.buffer.seek(0)
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file);self.flush = self.buffer.flush;self.writable = self.buffer.writable;self.write = lambda s: self.buffer.write(s.encode("ascii"));self.read = lambda: self.buffer.read().decode("ascii");self.readline = lambda: self.buffer.readline().decode("ascii")
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
  except:pass
  ii1=lambda:int(sys.stdin.readline().strip()) # for interger
  is1=lambda:sys.stdin.readline().strip() # for str
  iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
  isa=lambda:sys.stdin.readline().strip().split() # for List[str]
  mod=int(1e9 + 7);
  # from math import *
  # from math import inf 
  # from collections import *;
  from collections import defaultdict as dd
  # from collections import deque as dq 
  # from string import ascii_lowercase,ascii_uppercase
  # from functools import lru_cache
  # sys.setrecursionlimit(500000)
  # from bisect import bisect_left as bl,bisect_right as br
  # from heapq import heapify,heappop,heappush
  # def google(p):print('Case #'+str(p)+': ',end='')
  ###################### Start Here ######################
  from math import factorial
  def ncr(N):
      return N*(N-1)*(N-2)//6
  def fact(N):
      return factorial(N)
  for _ in range(ii1()):
      n = ii1()
      arr = []
      dic = dd(int)
      dic2 = dd(int)
      for _ in range(n):
          s,d = iia()
          arr.append((s,d))
          dic[s]+=1
          dic2[d]+=1
      ans1 = ncr(n)
      for s,d in arr:
          ans1 -= (dic[s]-1)*(dic2[d]-1)
      print(ans1)