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 FastScanner fs = new FastScanner();
  	static Scanner scn = new Scanner(System.in);
  	static PrintWriter pw = new PrintWriter(System.out);
  	static StringBuilder sb = new StringBuilder("");
  	static int[] BIT;
  	
  	public static void main(String[] args) {
  		int t = fs.nextInt();
  		for (int tt = 0; tt < t; tt++) {
  		int n = fs.nextInt();
  		int[] a = new int[n];
  		BIT = new int[n+1];
  		for (int i = 0; i < n; i++) {
  			a[i] = fs.nextInt();
  			a[i]--;
  		}
  		int sum = 0;
  		for (int i = 0; i < n; i++) {
  			sum += query(a[i] + 1, n);
  			add(a[i] + 1, 1);
  		}
  		if (sum % 2 == 0) sb.append("YES\n");
  		else {
  			sort(a);
  			boolean ok = false;
  			for (int i = 0; i < n - 1; i++) {
  				if (a[i] == a[i+1]) ok = true;
  			}
  		if (ok) sb.append("YES\n");
  		else sb.append("NO\n");
  		}
  		}
  		pw.print(sb.toString());
  		pw.close();
  	}
  	
  	static void sort(int[] a) {
  		ArrayList<Integer> al = new ArrayList<>();
  		for (int i : a)
  			al.add(i);
  		Collections.sort(al);
  		for (int i = 0; i < a.length; i++)
  			a[i] = al.get(i);
  	}
  	
  	static void add(int x, int k) {
  		for (int i = x; i < BIT.length; i+=i&-i) BIT[i] += k;
  	}
  	
  	static int query(int x) {
  		int res = 0;
  		for (int i = x; i > 0; i-=i&-i) res += BIT[i];
  		return res;
  	}
  	
  	static int query(int l, int r) {
  		return query(r) - query(l);
  	}
   
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		String next() {while (!st.hasMoreTokens())try {st = new StringTokenizer(br.readLine());} catch (IOException e) {}return st.nextToken();}
  		int nextInt() {return Integer.parseInt(next());}
  		long nextLong() {return Long.parseLong(next());}
  		double nextDouble() {return Double.parseDouble(next());}
  		int[] readArray(int n) { int[] a = new int[n]; for (int i = 0; i < n; i++) {a[i] = nextInt();} return a;}
  	}
  }