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
  import io, os
  #input = sys.stdin.buffer.readline
  input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  
  import itertools
  from collections import Counter, defaultdict
  
  t = int(input())
  for _ in range(t):
      n = int(input())
      A = [0]*n
      B = [0]*n
      AB = []
      ea = [set() for i in range(n)]
      eb = [set() for i in range(n)]
      for i in range(n):
          a, b = map(int, input().split())
          a, b = a-1, b-1
          A[i] = a
          B[i] = b
          AB.append((a, b))
          ea[a].add(b)
          eb[b].add(a)
  
      ans1 = 0
      da = [0]*n
      for a in A:
          da[a] += 1
      s1 = 0
      m1 = 0
      for i, c in enumerate(da):
          ans1 += m1*c
          m1 = s1*c+m1
          s1 += c
      #print(ans1)
      ans2 = 0
      db = [0]*n
      for b in B:
          db[b] += 1
      s2 = 0
      m2 = 0
      for i, c in enumerate(db):
          ans2 += m2*c
          m2 = s2*c+m2
          s2 += c
      #print(ans2)
      #AB.sort(key=lambda x: (x[0], x[1]))
  
      ans3 = n*(n-1)*(n-2)//6
      #print(ans3)
      #3
      for i in range(n):
          c = len(ea[i])
          if c >= 3:
              ans3 -= c*(c-1)*(c-2)//6
      for i in range(n):
          c = len(eb[i])
          if c >= 3:
              ans3 -= c*(c-1)*(c-2)//6
      #2
      for i in range(n):
          c = len(ea[i])
          if c < 2:
              continue
          temp = (c*(c-1)//2)*(n+2-c)
          s = 0
          for x in ea[i]:
              s += db[x]
          temp -= (c-1)*s
          #print('a', i, temp)
          #for x in ea[i]:
              #if i in eb[x]:
                  #eb[x].remove(i)
          ans3 -= temp
  
      for i in range(n):
          c = len(eb[i])
          if c < 2:
              continue
          temp = (c*(c-1)//2)*(n+2-c)
          s = 0
          for x in eb[i]:
              s += da[x]
          temp -= (c-1)*s
          #print('b', i, temp)
          ans3 -= temp
      for a, b in AB:
          temp = (da[a]-1)*(db[b]-1)
          ans3 -= temp
      #print(ans1, ans2, ans3)
      ans = ans1+ans2-ans3
      print(ans)
  