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.
from collections import defaultdict as dd
import sys

input = sys.stdin.readline



def dp_s(a):
    n = len(a)
    dp = [0]*n
    c = a[0]
    for i in range(1, n):
        dp[i] = dp[i - 1] + c*a[i]
        c += a[i]
    return dp


def dp_s2(a):
    dx = dd(int)
    dy = dd(int)
    n = len(a)
    dp = [0]*n
    dx[a[0][0]] += 1
    dy[a[0][1]] += 1
    for i in range(1, n):
        dp[i] += dp[i - 1]
        tot = i - dx[mat[i][0]] - dy[mat[i][1]]
        dp[i] += tot
    return dp


for _ in range(int(input())):
    n = int(input())
    mat = []
    dec_x = dd(int)
    dec_y = dd(int)
    for i in range(n):
        a, b = map(int, input().split())
        mat.append([a, b])
        dec_x[a] += 1
        dec_y[b] += 1
    xs = list(dec_x.values())
    ys = list(dec_y.values())
    dp1 = dp_s(xs)
    dp2 = dp_s(ys)
    ans1 = 0
    for i in range(1, len(dp1)):
        ans1 += xs[i]*dp1[i - 1]
    for i in range(1, len(dp2)):
        ans1 += ys[i]*dp2[i - 1]
    dp_m = dp_s2(mat)
    m = 0
    for i in range(2, n):
        m += dp_m[i - 1]
    print(ans1 - m)


