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 collections import Counter


def cn3(m):
    return m * (m - 1) * (m - 2) // 6


def solve_case():
    n = int(input())
    a = [-1] * n
    b = [-1] * n
    for i in range(n):
        a[i], b[i] = map(int, input().split())
    ca = Counter(a)
    cb = Counter(b)
    # sum all C(n, 3) over equals groups in a/b
    # e.g. if a = [ 4, 4, 4, 5, 5, 6 ]
    # then sum = cn3(3) + cn3(2) + cn3(1)
    ca_cn3_sum = sum(map(cn3, ca.values()))
    cb_cn3_sum = sum(map(cn3, cb.values()))
    # Triplets like:
    # a = [1, 1, 3]
    # b = [2, 4, 2]
    # both a & b have two equal items
    pair_sum = 0
    for i in range(n):
        pair_sum += (ca[a[i]] - 1) * (cb[b[i]] - 1)
    # they don't intersect due to task's conditions
    ans = cn3(n) - ca_cn3_sum - cb_cn3_sum \
        - pair_sum
    print(ans)


T = int(input())
for t in range(T):
    solve_case()
