human
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.
Solve the task in PYTHON3.

=====
What is a Python code to determine the number of ways to select three problems for the problemset?
assistant
#Here is Python code to determine the number of ways to select three problems for the problemset.

import sys

IS_INTERACTIVE = False
input = input  # type: ignore

# input: function
if not IS_INTERACTIVE:
    *data, = sys.stdin.read().split("\n")[::-1]

    def input():  # type: ignore
        return data.pop()

def fprint(*args, **kwargs):
    print(*args, **kwargs, flush=True)

def eprint(*args, **kwargs):
    print(*args, **kwargs, file=sys.stderr)

'''
1 1
2 1
2 2
2 4
3 4

1-2-5
1-3-5
1-4-5


'''


def naive(topics, diffs):
    n = len(topics)
    cnt = 0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if len({topics[i], topics[j], topics[k]}) == 3 or len({diffs[i], diffs[j], diffs[k]}) == 3:
                    cnt += 1
    return cnt

def fast(topics, diffs):
    tcnt = [0] * (n + 1)
    dcnt = [0] * (n + 1)
    for t, d in zip(topics, diffs):
        tcnt[t] += 1
        dcnt[d] += 1
    pairs = list(zip(topics, diffs))

    ans = n * (n - 1) * (n - 2) // 6
    for i in range(n):
        c0 = tcnt[topics[i]]
        c1 = dcnt[diffs[i]]
        c0 -= 1
        c1 -= 1
        if c0 and c1:
            ans -= c0 * c1

    return ans

'''
    n! / (n - 3)! / (3!)

    5 * 4 * 3

    1 2
    1 3
    2 2
'''

for _ in range(int(input())):
    n = int(input())
    topics = []
    diffs = []
    for _ in range(n):
        t, d = map(int, input().split())
        topics.append(t)
        diffs.append(d)

    print(fast(topics, diffs))