human
What is a Python code to determine the number of ways to select three problems for the problemset given the following:
PYTHON3 solution for "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.":
assistant
#Here is Python code to determine the number of ways to select three problems for the problemset.
 def naiveSolve():
    
    
    
    return


def solve():
    
    
    
    return


def main():
    
    t = int(input())
    allans = []
    for _ in range(t):
        n = int(input())
        a = []; b = []
        aCnt = [0] * (n + 1); bCnt = [0] * (n + 1)
        for __ in range(n):
            aa, bb = readIntArr()
            a.append(aa); b.append(bb)
            aCnt[aa] += 1; bCnt[bb] += 1
        ans = (n * (n - 1) * (n - 2)) // 6
        # print('initial ans:{}'.format(ans))
        badCnts = 0
        for i in range(n):
            badCnts += (aCnt[a[i]] - 1) * (bCnt[b[i]] - 1)
            # print('i:{} change:{}'.format(i, (aCnt[a[i]] - 1) * (bCnt[b[i]] - 1)))
        # badCnts //= 3
        ans -= badCnts
        allans.append(ans)
            
    multiLineArrayPrint(allans)
    
    return

import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.

def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]
 
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
    dv=defaultValFactory;da=dimensionArr
    if len(da)==1:return [dv() for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(x,y):
    print('? {} {}'.format(x,y))
    sys.stdout.flush()
    return int(input())
 
def answerInteractive(i1,j1,i2,j2):
    print('! {} {} {} {}'.format(i1,j1,i2,j2))
    sys.stdout.flush()
 
inf=float('inf')
# MOD=10**9+7
# MOD=998244353

from math import gcd,floor,ceil
# from math import floor,ceil # for Python2
 
for _abc in range(1):
    main()