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.
# cook your dish here
import sys
import bisect
from bisect import bisect_left as lb
from bisect import bisect_right as rb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**7)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs2(root,par):
    if par!=-1:
        dp[root]=dp[par]+1
    for i in range(1,20):
        if kthpar[root][i-1]!=-1:
            kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
    for child in gp[root]:
        if child==par:continue
        kthpar[child][0]=root
        dfs(child,root)
        
ans=0
a=[]
n,k=0,0
b=[]
vis=[]
tot=0
time=[]
time1=[]
adj=[]
mx=-1
eps=0.0000001
x=0
pref=[]
lvl=[]
#Fenwick Tree point update range sum
def update_it(bit,i,val):
    n=len(bit)
    while(i<n):
        bit[i]+=val
        i+=i&(-i)
def get_ans(bit,i):
    n=len(bit)
    tot=0
    while(i>0):
        tot+=bit[i]
        i-=i&(-i)
    return tot
def hnbhai(tc):
    n=sb()
    a=[]
    b=[]
    x={}
    y={}
    for i in range(n):
        t=sd()
        a.append(t[0])
        b.append(t[1])
        x[a[-1]]=x.get(a[-1],0)+1
        y[b[-1]]=y.get(b[-1],0)+1
    tot=(n*(n-1)*(n-2))//6
    for i in range(n):
        tot-=(x[a[i]]-1)*(y[b[i]]-1)
    print(tot)
for _ in range(sb()):
    hnbhai(_+1)
