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: class BIT:
  
      def __init__(self, L):
          self.N = len(L)
          self.bit = [0]*self.N
          for i,l in enumerate(L):
              self.add(i,l)
          self.N0 = 1
          while self.N0*2 <= self.N:
              self.N0 *= 2
  
      def add(self, a, w):
          x = a + 1
          for i in range(1000):
              self.bit[x-1] += w
              x += x & -x
              if x > self.N:
                  break
  
      def sum(self, a):
          x = a+1
          ret = 0
          for i in range(1000):
              ret += self.bit[x-1]
              x -= x & -x
              if x <= 0:
                  break
          return ret
  
      #you can use this function when the BIT has only non-negative values.
      def lower_bound(self, w):
          if w<=0:
              return 0
          x = 0
          k = self.N0
          while k>0:
              if x+k<=self.N:
                  if self.bit[x+k-1]<w:
                      w-=self.bit[x+k-1]
                      x+=k
              k//=2
          return x+1
  
  def solve():
      N = int(input())
      A = list(map(int,input().split()))
      if len(set(A))<len(A):
          print("YES")
          return
      bit = BIT([0]*N)
      A = [(a,i) for i,a in enumerate(A)]
      A.sort(key=lambda x:x[0], reverse=True)
      tento = 0
      buf = []
      prev = A[0][0]
      for a,i in A:
          if prev != a:
              for b in buf:
                  bit.add(b, 1)
              buf = []
              prev = a
          tento += bit.sum(i)
          buf.append(i)
      if tento%2==0:
          print("YES")
      else:
          print("NO")
  
  T = int(input())
  
  for i in range(T):
      solve()