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.
''' Interesting Pairs
https://codeforces.com/contest/1592/problem/A
'''

import io, os, sys
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline  # decode().strip() if str
output = sys.stdout.write

def solve(N, edges):
	row_cnt, col_cnt = {}, {}
	for r, c in edges:
		row_cnt[r] = row_cnt.get(r, 0) + 1
		col_cnt[c] = col_cnt.get(c, 0) + 1
	
	res = N*(N-1)*(N-2) // 6
	for r, c in edges:
		res -= (row_cnt[r] - 1) * (col_cnt[c] - 1)

	return res


def main():
	T = int(input())
	for _ in range(T):
		N = int(input())
		edges = []
		for _ in range(N):
			a, b = list(map(int, input().split()))
			edges.append((a-1, b-1))
		out = solve(N, edges)
		output(str(out) + '\n')


if __name__ == '__main__':
	main()

