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 math;import heapq;import string;from collections import deque;from bisect import bisect,bisect_left,bisect_right,insort;import sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());H=1000000000+7
  for _ in range(I()):
  	n=I()
  	l=[]
  	for i in range(n):
  		l.append(L())
  	d1={};d2={}
  	for i in range(n):
  		d1[l[i][0]]=d1.get(l[i][0],0)+1
  		d2[l[i][1]]=d2.get(l[i][1],0)+1
  	p=0
  	for i in range(n):
  		p+=(d1[l[i][0]]-1)*(d2[l[i][1]]-1)
  	print(n*(n-1)*(n-2)//6  -  p)