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.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.Arrays;
  import java.util.HashSet;
  import java.util.StringTokenizer;
  import java.util.TreeSet;
  
  public class D {
  	public static void main(String args[]) throws Exception {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(br.readLine().trim());
  		StringBuilder sb = new StringBuilder();
  		while (t-- > 0) {
  			int N = Integer.parseInt(br.readLine().trim());
  			int arr[] = nextIntArray(N, br);
  			sb.append(solve(arr)).append("\n");
  		}
  		br.close();
  		System.out.println(sb);
  	}
  
  	private static String solve(int arr[]) {
  		HashSet<Integer> hs = new HashSet<>();
  		for (int i = 0; i < arr.length; i++)
  			hs.add(arr[i]);
  
  		if (hs.size() < arr.length)
  			return "YES";
  
  		long inv = mergeSortAndCount(arr, 0, arr.length - 1);
  		// System.out.println(inv);
  		if (inv % 2 == 0)
  			return "YES";
  		return "NO";
  
  	}
  
  	private static int mergeAndCount(int[] arr, int l,
  			int m, int r) {
  
  		// Left subarray
  		int[] left = Arrays.copyOfRange(arr, l, m + 1);
  
  		// Right subarray
  		int[] right = Arrays.copyOfRange(arr, m + 1, r + 1);
  
  		int i = 0, j = 0, k = l, swaps = 0;
  
  		while (i < left.length && j < right.length) {
  			if (left[i] <= right[j])
  				arr[k++] = left[i++];
  			else {
  				arr[k++] = right[j++];
  				swaps += (m + 1) - (l + i);
  			}
  		}
  		while (i < left.length)
  			arr[k++] = left[i++];
  		while (j < right.length)
  			arr[k++] = right[j++];
  		return swaps;
  	}
  
  	// Merge sort function
  	private static int mergeSortAndCount(int[] arr, int l,
  			int r) {
  
  		// Keeps track of the inversion count at a
  		// particular node of the recursion tree
  		int count = 0;
  
  		if (l < r) {
  			int m = (l + r) / 2;
  
  			// Total inversion count = left subarray count
  			// + right subarray count + merge count
  
  			// Left subarray count
  			count += mergeSortAndCount(arr, l, m);
  
  			// Right subarray count
  			count += mergeSortAndCount(arr, m + 1, r);
  
  			// Merge count
  			count += mergeAndCount(arr, l, m, r);
  		}
  
  		return count;
  	}
  
  	private static int[] nextIntArray(int N, BufferedReader br) throws Exception {
  		StringTokenizer st = new StringTokenizer(br.readLine().trim(), " ");
  		int arr[] = new int[N];
  		for (int i = 0; i < N; i++)
  			arr[i] = Integer.parseInt(st.nextToken());
  		return arr;
  	}
  }
  