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: from collections import Counter
  from queue import Queue
  import math
  
  
  def minSwaps(arr):
      n = len(arr)
  
      # Create two arrays and use
      # as pairs where first array
      # is element and second array
      # is position of first element
      arrpos = [*enumerate(arr)]
  
      # Sort the array by array element
      # values to get right position of
      # every element as the elements
      # of second array.
      arrpos.sort(key=lambda it: it[1])
  
      # To keep track of visited elements.
      # Initialize all elements as not
      # visited or false.
      vis = {k: False for k in range(n)}
  
      # Initialize result
      ans = 0
      for i in range(n):
  
          # alreadt swapped or
          # alreadt present at
          # correct position
          if vis[i] or arrpos[i][0] == i:
              continue
  
          # find number of nodes
          # in this cycle and
          # add it to ans
          cycle_size = 0
          j = i
  
          while not vis[j]:
              # mark node as visited
              vis[j] = True
  
              # move to next node
              j = arrpos[j][0]
              cycle_size += 1
  
          # update answer by adding
          # current cycle
          if cycle_size > 0:
              ans += (cycle_size - 1)
  
      # return answer
      return ans
  
  if __name__ == '__main__':
      '''
      Test case:
  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
  
      '''
      T = int(input())
      for t in range(T):
          n = int(input())
          A = [int(s) - 1 for s in input().split(' ')]
          d = {}
          ans = ""
          for a in A:
              if a in d:
                  ans = "YES"
                  break
              else:
                  d[a] = 0
          if ans == "":
              ans = "YES" if minSwaps(A) % 2 == 0 else "NO"
          print(ans)