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.
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python program to determine the number of ways to select three problems for the problemset.

import os
import sys
from collections import Counter, defaultdict, deque
from io import BytesIO, IOBase
from math import gcd, inf, sqrt, ceil
def lcm(a, b):
    return (a * b) // gcd(a, b)
'''
    mod = 10 ** 9 + 7
    fac = [1]
    for i in range(1, 2 * 10 ** 5 + 1):
        fac.append((fac[-1] * i) % mod)
    fac_in = [pow(fac[-1], mod - 2, mod)]
    for i in range(2 * 10 ** 5, 0, -1):
        fac_in.append((fac_in[-1] * i) % mod)
    fac_in.reverse()
    def comb(a, b):
        if a < b:
            return 0
        return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''
#MAXN = 10000004
# spf = [0 for i in range(MAXN)]
# adj = [[] for i in range(MAXN)]
def sieve():
    global spf, adj, MAXN
    spf[1] = 1
    for i in range(2, MAXN):
        spf[i] = i

    for i in range(2, MAXN):
        if i * i > MAXN:
            break
        if (spf[i] == i):
            for j in range(i * i, MAXN, i):
                if (spf[j] == j):
                    spf[j] = i
def getdistinctFactorization(n):
    global adj, spf, MAXN
    for i in range(1, n + 1):
        index = 1
        x = i
        if (x != 1):
            adj[i].append(spf[x])
        x = x // spf[x]
        while (x != 1):
            if (adj[i][index - 1] != spf[x]):
                adj[i].append(spf[x])
                index += 1
            x = x // spf[x]
def printDivisors(n):
    i = 2
    z = [1, n]
    while i <= sqrt(n):
        if (n % i == 0):
            if (n / i == i):
                z.append(i)
            else:
                z.append(i)
                z.append(n // i)
        i = i + 1
    return z
def create(n, x, f):
    pq = len(bin(n)[2:])
    if f == 0:
        tt = min
    else:
        tt = max
    dp = [[inf] * n for _ in range(pq)]
    dp[0] = x
    for i in range(1, pq):
        for j in range(n - (1 << i) + 1):
            dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
    return dp
def enquiry(l, r, dp, f):
    if l > r:
        return inf if not f else -inf
    if f == 1:
        tt = max
    else:
        tt...