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: from sys import stdin,stdout
  from math import gcd,sqrt,factorial,pi,inf
  from collections import deque,defaultdict
  from bisect import bisect,bisect_left
  from time import time
  from itertools import permutations as per
  from heapq import heapify,heappush,heappop,heappushpop
  input=stdin.readline
  R=lambda:map(int,input().split())
  I=lambda:int(input())
  S=lambda:input().rstrip('\r\n')
  L=lambda:list(R())
  P=lambda x:stdout.write(str(x)+'\n')
  lcm=lambda x,y:(x*y)//gcd(x,y)
  nCr=lambda x,y:(f[x]*inv((f[y]*f[x-y])%N))%N
  inv=lambda x:pow(x,N-2,N)
  sm=lambda x:(x**2+x)//2
  N=10**9+7
  
  for _ in range(I()):
      n=I()
      a=[0]*(n+1)
      b=[0]*(n+1)
      d=[]
      for i in range(n):
          x,y=R()
          a[x]+=1
          b[y]+=1
          d+=(x,y),
      ans=(n*(n-1)*(n-2))//6
      for i in range(n):
          ans-=(a[d[i][0]]-1)*(b[d[i][1]]-1)
      print(ans)
  
  
  
  
  
  
  