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 io
import os

input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

from collections import defaultdict
from itertools import combinations

for _ in range(int(input())):
    n = int(input())
    l_d = defaultdict(int)
    r_d = defaultdict(int)
    l = set()
    r = set()
    both = set()
    for ___ in range(n):
        a, b = map(int, input().split())
        l_d[a] += 1
        r_d[b] += 1
        both.add((a, b))
        l.add(a)
        r.add(b)

    l_d_copy = l_d.copy()
    r_d_copy = r_d.copy()
    deleted = []
    for pair in both:
        if l_d[pair[0]] > 1 and r_d[pair[1]] > 1:
            l_d[pair[0]] -= 1
            r_d[pair[1]] -= 1
            deleted.append(pair)

    for pair in deleted:
        both.remove(pair)

    deleted = []
    for pair in both:
        if l_d[pair[0]] > 1:
            l_d[pair[0]] -= 1
            deleted.append(pair)
        elif r_d[pair[1]] > 1:
            r_d[pair[1]] -= 1
            deleted.append(pair)

    for pair in deleted:
        both.remove(pair)

    # print(both)

    left = len(list(combinations(l, 3)))
    right = len(list(combinations(r, 3)))
    # print(left, right)
    ans = left + right
    for num in l_d_copy:
        ans += left * (l_d_copy[num] - 1)

    for num in r_d_copy:
        ans += right * (r_d_copy[num] - 1)

    print(ans - len(list(combinations(both, 3))))



