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
input = sys.stdin.buffer.readline 

def process(A):
    n = len(A)
    total = n*(n-1)*(n-2)//6
    pairs = {}
    topic = {}
    diff = {}
    for i in range(n):
        a, b = A[i]
        if (a, b) not in pairs:
            pairs[(a, b)] = 0
        pairs[(a, b)]+=1
        if a not in topic:
            topic[a] = 0
        topic[a]+=1
        if b not in diff:
            diff[b] = 0
        diff[b]+=1
    bad = 0
    for x in pairs:
        a, b = x
        y = pairs[x]
        z1 = topic[a]
        z2 = diff[b]
        #one topic, one difficulty
        bad_triple1 = y*(y-1)*(y-2)//6
      #  print(bad_triple, 'One')
        bad+=bad_triple1
        
        #one topic, two difficulties
        bad_triple2 = (y*(y-1)//2)*(z1-2)-y*(y-1)*(y-2)//6
      #  print(bad_triple, 'Two')
        bad+=bad_triple2
        
        #two topics, one difficulty
        bad_triple3 = (y*(y-1)//2)*(z2-2)-y*(y-1)*(y-2)//6
      #  print(bad_triple, "Three")
        bad+=bad_triple3
    #two topics, two difficulties
    #case 1
    #a x
    #a x
    #b y
        unmatched_pairs = (n-z1-z2+y)
        bad_triple4 = ((y*(y-1))//2)*unmatched_pairs
   #     print(bad_triple, 'Four')
        bad+=bad_triple4
        #two topics, two difficulties
        #case 2
        #a x
        #a y
        #b x  
        other_a = (z1-y)
        other_b = (z2-y)
        bad_triple5 = y*other_a*other_b
     #   print(bad_triple, "Five")
        bad+=bad_triple5
    return total-bad

t = int(input())
for i in range(t):
    n = int(input())
    A = []
    for j in range(n):
        a, b = [int(x) for x in input().split()]
        A.append([a, b])
    answer = process(A)
    sys.stdout.write(str(answer)+'\n')