description: Petya has an array of integers a_1, a_2, …, a_n. He only likes sorted arrays. Unfortunately, the given array could be arbitrary, so Petya wants to sort it.
  
  Petya likes to challenge himself, so he wants to sort array using only 3-cycles. More formally, in one operation he can pick 3 pairwise distinct indices i, j, and k (1 ≤ i, j, k ≤ n) and apply i → j → k → i cycle to the array a. It simultaneously places a_i on position j, a_j on position k, and a_k on position i, without changing any other element.
  
  For example, if a is [10, 50, 20, 30, 40, 60] and he chooses i = 2, j = 1, k = 5, then the array becomes [\underline{50}, \underline{40}, 20, 30, \underline{10}, 60].
  
  Petya can apply arbitrary number of 3-cycles (possibly, zero). You are to determine if Petya can sort his array a, i. e. make it non-decreasing.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5 ⋅ 10^5). Description of the test cases follows.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the length of the array a.
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n).
  
  It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5.
  
  Output
  
  For each test case, print "YES" (without quotes) if Petya can sort the array a using 3-cycles, and "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower).
  
  Example
  
  Input
  
  
  7
  1
  1
  2
  2 2
  2
  2 1
  3
  1 2 3
  3
  2 1 3
  3
  3 1 2
  4
  2 1 4 3
  
  
  Output
  
  
  YES
  YES
  NO
  YES
  NO
  YES
  YES
  
  Note
  
  In the 6-th test case Petya can use the 3-cycle 1 → 3 → 2 → 1 to sort the array.
  
  In the 7-th test case Petya can apply 1 → 3 → 2 → 1 and make a = [1, 4, 2, 3]. Then he can apply 2 → 4 → 3 → 2 and finally sort the array.
solution: #!/usr/bin/env python3
  # from typing import *
  
  import sys
  import io
  import math
  import collections
  import decimal
  import itertools
  import bisect
  import heapq
  
  
  def input():
      return sys.stdin.readline()[:-1]
  
  
  # sys.setrecursionlimit(1000000)
  
  # _INPUT = """7
  # 1
  # 1
  # 2
  # 2 2
  # 2
  # 2 1
  # 3
  # 1 2 3
  # 3
  # 2 1 3
  # 3
  # 3 1 2
  # 4
  # 2 1 4 3
  
  # """
  # sys.stdin = io.StringIO(_INPUT)
  
  INF = 10**10
  
  YES = 'YES'
  NO = 'NO'
  
  def solve(N, A):
      if N == 1:
          return 'YES'
  
      counter = collections.Counter(A)
      if counter.most_common(1)[0][1] > 1:
          return 'YES'
      
      D = {a: i for i, a in enumerate(sorted(set(A)))}
      B = [D[a] for a in A]
      Pos = {B[i]: i for i in range(N)}
  
      for i in range(N-2):
          p = Pos[i]
          if p == i:
              continue
          elif p == i+1:
              k1, k2, k3 = i, i+1, i+2
              Pos[B[k1]], Pos[B[k2]], Pos[B[k3]] = k3, k1, k2
              B[k1], B[k2], B[k3] = B[k2], B[k3], B[k1]
          else:
              k1, k2, k3 = i, i+1, p
              Pos[B[k1]], Pos[B[k2]], Pos[B[k3]] = k2, k3, k1
              B[k1], B[k2], B[k3] = B[k3], B[k1], B[k2]
      
      if B[N-2] == N-2 and B[N-1] == N-1:
          return 'YES'
  
      return 'NO'
  
  T0 = int(input())
  for _ in range(T0):
      N = int(input())
      A = list(map(int, input().split()))
      print(solve(N, A))