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)
    g = HopcroftKarp(n, n)
    for _ in [0] * n:
        a, b = li0input()
        A[a] += 1
        B[b] += 1
        g.add_edge(a, b)
    g.flow()
    joint = g.matching().count(1)
    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():
        ans1 -= (n - v) * v * (v - 1) // 2
        ans1 -= v * (v - 1) * (v - 2) // 6
    for v in B.values():
        ans2 -= (n - v) * v * (v - 1) // 2
        ans2 -= v * (v - 1) * (v - 2) // 6
    print(ans1 + ans2 - joint)