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
  from math import log2, ceil, sqrt, gcd
  from _collections import deque
  import heapq as hp
  from bisect import bisect_left, bisect_right
  from math import cos, sin
  from itertools import permutations
  from operator import itemgetter
  
  # sys.setrecursionlimit(2*10**5+10000)
  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)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  mod = 10 ** 9 + 7
  
  
  def solve(x):
      d = dict()
      for i in b[x]:
          d[i] = d.get(i, 0) + 1
      ck = [d[i] for i in d]
      if len(ck) < 3:
          return 0
      m = len(ck)
      dp = [[0] * m for _ in range(3)]
      dp[0][0] = ck[0]
      for i in range(1, len(ck)):
          dp[0][i] = dp[0][i - 1] + ck[i]
      for i in range(1, 3):
          for j in range(i, m):
              dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1] * ck[j]
      print(dp[-1][-1], '->')
      return dp[-1][-1]
  
  
  for _ in range(int(input())):
      n = int(input())
      a = [[] for _ in range(n + 1)]
      b = [[] for _ in range(n + 1)]
      d = dict()
      for _ in range(n):
          x, y = map(int, input().split())
          a[x].append(y)
          b[y].append(x)
          d[y] = d.get(y,0) + 1
      ans = 0
      for i in range(1, n + 1):
          t = len(a[i])
          ct = 0
          for i in a[i]:
              ct += d[i] - 1
          ans += ct*max((t-1),0)
      # print(ans)
      print((n * (n - 1) * (n - 2)) // 6 - ans)
  