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.
#!/usr/bin/env python3
# from typing import *

import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq


def input():
    return sys.stdin.readline()[:-1]


# sys.setrecursionlimit(1000000)

# _INPUT = """2
# 4
# 2 4
# 3 4
# 2 1
# 1 3
# 5
# 1 5
# 2 4
# 3 3
# 4 2
# 5 1

# """
# sys.stdin = io.StringIO(_INPUT)

INF = 10**10

def solve(N, T):
    counter1 = collections.Counter([a for a, b in T])
    counter2 = collections.Counter([b for a, b in T])

    v1 = 0
    for a, n in counter1.items():
        if n >= 2:
            v1 += n*(n-1)//2 * (N-n)

    v2 = 0
    for a, n in counter1.items():
        if n >= 3:
            v2 += n*(n-1)*(n-2)//6

    v3 = 0
    for a, b in T:
        v3 += (counter1[a]-1) * (counter2[b]-1)
    
    v4 = 0
    for a, n in counter2.items():
        if n >= 2:
            v4 += n*(n-1)//2 * (N-n)

    v5 = 0
    for a, n in counter2.items():
        if n >= 3:
            v5 += n*(n-1)*(n-2)//6

    m = N*(N-1)*(N-2)//6
    ans = (m-v1-v2) + (m-v4-v5) - (m-v1-v2+v3-v4-v5)
    return ans


T0 = int(input())
for _ in range(T0):
    N = int(input())
    T = []
    for _ in range(N):
        a, b = map(int, input().split())
        T.append((a, b))
    print(solve(N, T))
