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.
import sys
import math
from bisect import bisect_left
import heapq
from collections import deque
from itertools import product, permutations
import random
def II():
	return int(sys.stdin.readline())
 
def LI():
	return list(map(int, sys.stdin.readline().split()))
 
def MI():
	return map(int, sys.stdin.readline().split())
 
def SI():
	return sys.stdin.readline().strip()
 
def C(n, k, mod):
    return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
 
def lcm(a,b):
    return abs(a*b) // math.gcd(a, b)

def helper(n):
	if n >=3:
		return (n*(n-1)*(n-2))//6
	return 0

def helper2(n):
	if n>1:
		return (n*(n-1))//2
	return 0
 
mod = 10**9+7

for _ in range(II()):
	n = II()
	d = {} #сложности
	d1 = {} #задачи
	d2 = {}
	ans = helper(n)
	for i in range(n):
		a,b = MI()
		if d.get(b):
			d[b]+=1
			d2[b].append(a)
		else:
			d[b] = 1
			d2[b] = [a]
		if d1.get(a):
			d1[a]+=1
		else:
			d1[a] = 1
	
	s = 0
	s1 = 0
	for i in d:
		s+=helper2(d[i])
	for i in d1:
		s1+=helper2(d1[i])
	print(max(ans-s*s1,0))