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
from sys import stdin

tt = int(stdin.readline())

ANS = []
for loop in range(tt):

    n = int(stdin.readline())

    ab = []
    anum = {}
    bnum = {}

    for i in range(n):
        a,b = map(int,stdin.readline().split())
        ab.append((a,b))

        if a not in anum:
            anum[a] = 0
        if b not in bnum:
            bnum[b] = 0

        anum[a] += 1
        bnum[b] += 1


    twonl = 0
    for a in anum:
        c = anum[a]
        twonl += c * (c-1) // 2 * (n-c)
    for b in bnum:
        c = bnum[b]
        twonl += c * (c-1) // 2 * (n-c)

    tans = 0
    for a,b in ab:
        tans += (anum[a]-1) * (bnum[b]-1)

    aans = 0
    for a in anum:
        c = anum[a]
        aans += c * (c-1) * (c-2) // 6
    for b in bnum:
        c = bnum[b]
        aans += c * (c-1) * (c-2) // 6

    #print (twonl , tans , ans)
    aans = n*(n-1)*(n-2)//6  - (twonl - tans + aans) #both


    #a only
    dpA = [0,0,0]
    for a in anum:
        c = anum[a]
        dpA[2] += dpA[1] * c
        dpA[1] += dpA[0] * c
        dpA[0] += c

    dpB = [0,0,0]
    for b in bnum:
        c = bnum[b]
        dpB[2] += dpB[1] * c
        dpB[1] += dpB[0] * c
        dpB[0] += c

    twoans = dpA[-1] + dpB[-1]
    ANS.append(str(twoans - aans))

print ("\n".join(ANS))