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 sys
import io, os
#input = sys.stdin.buffer.readline
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

import itertools
from collections import Counter, defaultdict

t = int(input())
for _ in range(t):
    n = int(input())
    A = [0]*n
    B = [0]*n
    AB = []
    ea = [set() for i in range(n)]
    eb = [set() for i in range(n)]
    for i in range(n):
        a, b = map(int, input().split())
        a, b = a-1, b-1
        A[i] = a
        B[i] = b
        AB.append((a, b))
        ea[a].add(b)
        eb[b].add(a)

    ans1 = 0
    da = [0]*n
    for a in A:
        da[a] += 1
    s1 = 0
    m1 = 0
    for i, c in enumerate(da):
        ans1 += m1*c
        m1 = s1*c+m1
        s1 += c
    #print(ans1)
    ans2 = 0
    db = [0]*n
    for b in B:
        db[b] += 1
    s2 = 0
    m2 = 0
    for i, c in enumerate(db):
        ans2 += m2*c
        m2 = s2*c+m2
        s2 += c
    #print(ans2)
    #AB.sort(key=lambda x: (x[0], x[1]))

    ans3 = n*(n-1)*(n-2)//6
    #print(ans3)
    #3
    for i in range(n):
        c = len(ea[i])
        if c >= 3:
            ans3 -= c*(c-1)*(c-2)//6
    for i in range(n):
        c = len(eb[i])
        if c >= 3:
            ans3 -= c*(c-1)*(c-2)//6
    #2
    for i in range(n):
        c = len(ea[i])
        if c < 2:
            continue
        temp = (c*(c-1)//2)*(n+2-c)
        s = 0
        for x in ea[i]:
            s += db[x]
        temp -= (c-1)*s
        #print('a', i, temp)
        #for x in ea[i]:
            #if i in eb[x]:
                #eb[x].remove(i)
        ans3 -= temp

    for i in range(n):
        c = len(eb[i])
        if c < 2:
            continue
        temp = (c*(c-1)//2)*(n+2-c)
        s = 0
        for x in eb[i]:
            s += da[x]
        temp -= (c-1)*s
        #print('b', i, temp)
        ans3 -= temp
    for a, b in AB:
        temp = (da[a]-1)*(db[b]-1)
        ans3 -= temp
    #print(ans1, ans2, ans3)
    ans = ans1+ans2-ans3
    print(ans)
