human
Can you solve the below in PYTHON3?
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.
assistant
#Here is Python program to determine the number of ways to select three problems for the problemset.

import sys
input = sys.stdin.readline
def iinput(): return int(input())
def sinput(): return input().rstrip()
def i0input(): return int(input()) - 1
def linput(): return list(input().split())
def liinput(): return list(map(int, input().split()))
def miinput(): return map(int, input().split())
def li0input(): return list(map(lambda x: int(x) - 1, input().split()))
def mi0input(): return map(lambda x: int(x) - 1, input().split())
INF = 10**20
MOD = 1000000007

# Hopcroft-Karp Algorithm
# from collections import deque
# class HopcroftKarp:
#     def __init__(self, N0, N1):
#         self.N0 = N0
#         self.N1 = N1
#         self.N = N = 2+N0+N1
#         self.G = [[] for i in range(N)]
#         for i in range(N0):
#             forward = [2+i, 1, None]
#             forward[2] = backward = [0, 0, forward]
#             self.G[0].append(forward)
#             self.G[2+i].append(backward)
#         self.backwards = bs = []
#         for i in range(N1):
#             forward = [1, 1, None]
#             forward[2] = backward = [2+N0+i, 0, forward]
#             bs.append(backward)
#             self.G[2+N0+i].append(forward)
#             self.G[1].append(backward)

#     def add_edge(self, fr, to):
#         #assert 0 <= fr < self.N0
#         #assert 0 <= to < self.N1
#         v0 = 2 + fr
#         v1 = 2 + self.N0 + to
#         forward = [v1, 1, None]
#         forward[2] = backward = [v0, 0, forward]
#         self.G[v0].append(forward)
#         self.G[v1].append(backward)

#     def bfs(self):
#         G = self.G
#         level = [None]*self.N
#         deq = deque([0])
#         level[0] = 0
#         while deq:
#             v = deq.popleft()
#             lv = level[v] + 1
#             for w, cap, _ in G[v]:
#                 if cap and level[w] is None:
#                     level[w] = lv
#                     deq.append(w)
#         self.level = level
#         return level[1] is not None

#     def dfs(self, v, t):
#         if v == t:
#             return 1
#         level = self.level
#     ...