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.
from math import *
#from bisect import *
from collections import *
#from random import *
#from decimal import *"""
#from heapq import *
#from itertools import *
import sys
input=sys.stdin.readline
def inp():
    return int(input())
def st():
    return input().rstrip('\n')
def lis():
    return list(map(int,input().split()))
def ma():
    return map(int,input().split())
t=inp()
while(t):
    t-=1
    n=inp()
    topic={}
    diff={}
    res=(n*(n-1)*(n-2))//6
    r=[]
    for i in range(n):
        a,b=ma()
        r.append([a,b])
        try:
            topic[a].append(b)
        except:
            topic[a]=[b]
        try:
            diff[b].append(a)
        except:
            diff[b]=[a]
    ex=0
    store={}
    vis=set()
    for i in range(n):
        z=r[i][0]
        if(z in vis):
            continue
        vis.add(z)
        if(len(topic[z])<=1):
            continue
        for j in range(len(topic[z])):
            curdiff=topic[z][j]
            if(len(diff[curdiff])<=1):
                continue
            fi=len(topic[z])-1
            pro=len(diff[curdiff])-1
            res-=(pro)*fi
    print(res)
            
            
        
        
    
        
