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: import java.util.*;
  
  /*
  Wonderful observations from GennadyTran:
  _ always can do shifitng to move last laargest n-2 to correct position. Then if we have 2 same number we can correct positions for first 2 one
  _Now all numbers are distinct to form permutaion, can convert to cycles notation. Each time to break cycle, we only can move 2 number to correct positions (51234)=(512).(534)
  len of cycle will reduce by2, last cycle must have len 3 => Odd cycle = even transposition will be able to reduce
  _ Even cycle: we can merge 2 even cycle. and convert into 2 odd cycle => even number of even cycles => yes , otherwise no
  _ Tags: transposition, compose and decompose
  */ 
  
  public class MyClass {
      static Scanner sc= new Scanner(System.in);
      static boolean[] vis;
      static int even,odd;
      public static void main(String args[]) {
        int t= sc.nextInt();
        while(t-->0){
            int n= sc.nextInt();
            vis= new boolean[n+1];
            even=0; odd=0;
            int[] a= new int[n+1];
            Set<Integer> num= new HashSet<>();
            boolean same= false;
            for(int i=1;i<=n;i++){
                a[i]= sc.nextInt();
                if(num.contains(a[i])) same= true;
                num.add(a[i]);
            }
            if(same){System.out.println("YES"); continue;}
            
            for(int i=1;i<=n;i++){
                if(vis[i]) continue;
                dfs(a,i,0);
            }
            if(even%2==1) System.out.println("NO");
            else System.out.println("YES");
            
        }
      }
      
      public static void dfs(int[] a, int i, int len){
          len++;
          if(i==a[i] || vis[a[i]]) {
              if(len%2==0) even++;
              else odd++;
              return;
          }
          vis[i]= true;
          dfs(a,a[i],len);
      }
  }