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: for _ in range(int(input())):
      n = int(input())
      a = list(map(int,input().split()))
      if n==1:
          print("YES")
          continue
  
      h = n != len(set(a))
  
      b = sorted(a)
      d = dict()
  
      for i,x in enumerate(a):
          d[x] = d.get(x,set())
          d[x].add(i)
  
      i = n-1
      while i>1:
          r = a[i]
          t = b[i]
          if r!=t:
              ti = 0
              for tix in d[t]:
                  ti = tix
                  break
  
              z = 0
              if ti==z:
                  z=1
              az = a[z]
  
              a[i] = t
              d[t].remove(ti)
              a[ti] = a[z]
              d[az].remove(z)
              d[az].add(ti)
              a[z] = r
              d[r].remove(i)
              d[r].add(z)
  
          i-=1
  
      if a[0]==b[0] and a[1]==b[1]:
          print("YES")
      else:
          if h:
              print("YES")
          else:
              print("NO")
  