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.*;
  import java.io.*;
  
  public class D {
  	
  	static int[] nums;
  	public static void main (String[] args) throws IOException {
  		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
  		int T = Integer.parseInt(in.readLine());
  		
  		for (int t = 0; t < T; t++) {
  			int n = Integer.parseInt(in.readLine());
  			HashSet<Integer> set = new HashSet<Integer>();
  			boolean dup = false;
  			nums = new int[n];
  			StringTokenizer line = new StringTokenizer(in.readLine());
  			for (int i = 0; i < n; i++) {
  				int current = Integer.parseInt(line.nextToken());
  				nums[i] = current;
  				if (set.contains(current))
  					dup = true;
  				else
  					set.add(current);
  			}
  			
  			int swaps = minSwaps(nums);
  			
  			if (dup || swaps % 2 == 0)
  				System.out.println("YES");
  			else
  				System.out.println("NO");
  		}
  		
  		
  	}
  	
  	//boonked from geeksforgeeks.org
  	public static int minSwaps(int[] nums)
      {
          int len = nums.length;
          HashMap<Integer, Integer> map = new HashMap<>();
          for(int i=0;i<len;i++)
              map.put(nums[i], i);
               
          Arrays.sort(nums);  
           
            // To keep track of visited elements. Initialize
          // all elements as not visited or false.
          boolean[] visited = new boolean[len];
          Arrays.fill(visited, false);
           
            // Initialize result
          int ans = 0;
          for(int i=0;i<len;i++) {
             
                // already swapped and corrected or
              // already present at correct pos
              if(visited[i] || map.get(nums[i]) == i)
                  continue;
                   
              int j = i, cycle_size = 0;
              while(!visited[j]) {
                  visited[j] = true;
                   
                    // move to next node
                  j = map.get(nums[j]);
                  cycle_size++;
              }
               
                // Update answer by adding current cycle.
              if(cycle_size > 0) {
                  ans += (cycle_size - 1);
              }
          }
          return ans;
      }
  }
  