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.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
#         for e in self.it[v]:
#             w, cap, rev = e
#             if cap and level[v] < level[w] and self.dfs(w, t):
#                 e[1] = 0
#                 rev[1] = 1
#                 return 1
#         return 0

#     def flow(self):
#         flow = 0
#         G = self.G
#         bfs = self.bfs; dfs = self.dfs
#         while bfs():
#             *self.it, = map(iter, G)
#             while dfs(0, 1):
#                 flow += 1
#         return flow

#     def matching(self):
#         return [cap for _, cap, _ in self.backwards]

t = iinput()
from collections import defaultdict
for _ in [0] * t:
    n = iinput()
    A = defaultdict(int)
    B = defaultdict(int)
    all = []
    # g = HopcroftKarp(n, n)
    for _ in [0] * n:
        a, b = li0input()
        A[a] += 1
        B[b] += 1
        all.append((a, b))
    #     g.add_edge(a, b)
    # joint = g.flow()
    # joint = joint * (joint - 1) * (joint - 2) // 6
    ans1 = n * (n - 1) * (n - 2) // 6
    ans2 = n * (n - 1) * (n - 2) // 6
    for v in A.values():
        if v >= 2:
            ans1 -= (n - v) * v * (v - 1) // 2
        if v >= 3:
            ans1 -= v * (v - 1) * (v - 2) // 6
    for v in B.values():
        if v >= 2:
            ans2 -= (n - v) * v * (v - 1) // 2
        if v >= 3:
            ans2 -= v * (v - 1) * (v - 2) // 6
    bar_joint = 0
    for a, b in all:
        bar_joint += max(0, A[a] - 1) * max(0, B[b] - 1)
    joint = ans1 + ans2 + bar_joint - n * (n - 1) * (n - 2) // 6
    print(ans1 + ans2 - joint)