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.IOException;
  import java.io.InputStreamReader;
  
  public class d {
  	public static void main(String[] args) throws IOException {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		String[] line; int $ = -1;
  		
  		
  		$=0; line=br.readLine().split(" ");
  		int T = Integer.parseInt(line[$++]);
  		while(T-->0) {
  			$=0; line=br.readLine().split(" ");
  			int N = Integer.parseInt(line[$++]);
  
  			$=0; line=br.readLine().split(" ");
  			int[] arr = new int[N];
  			for(int n=0;n<N;n++) {
  				arr[n] = Integer.parseInt(line[$++]);
  			}
  			if(N == 1) {
  				System.out.println("YES");
  				continue;
  			}
  			
  			ST.theArray = arr;
  			ST st = new ST(0,N-1);
  			
  			// of course if there are any duplicates whatsoever, then of course it's possible
  			// we'll check this at the end
  
  //			st.print(true);
  			
  			for(int n=0;n<N-2;n++) {
  				// find the smallest
  				int dex = st.queryMinDex(n, N-1);
  //				System.out.println("min is "+dex);
  				if(dex == n) {
  					continue;
  				}
  				
  				int a = N-1;
  				if(a == dex) a--;
  				
  				int nVal = arr[n];
  				int dexVal = arr[dex];
  				int aVal = arr[a];
  				
  				// this cycle
  				st.update(n, dexVal);
  //				st.print(true);
  				st.update(dex, aVal);
  //				st.print(true);
  				st.update(a, nVal);
  				
  				arr[n] = dexVal;
  				arr[dex] = aVal;
  				arr[a] = nVal;
  
  //				st.print(true);
  			}
  			
  			if(arr[N-2] > arr[N-1]) {
  				// one last hope
  				// are there ANY duplicate elements?
  				arr[N-2] ^= arr[N-1];
  				arr[N-1] ^= arr[N-2];
  				arr[N-2] ^= arr[N-1];
  				
  				boolean hasDupe = false;
  				for(int n = 1; n < N; n++) {
  					if(arr[n-1] == arr[n]) {
  						hasDupe = true;
  						break;
  					}
  				}
  				
  				if(hasDupe) {
  					System.out.println("YES");
  				}
  				else {
  					// i am disappoint.
  					System.out.println("NO");
  				}
  			}
  			else {
  				System.out.println("YES");
  			}
  			
  			// take the last three elements and sort them
  //			ArrayList<Integer> lastEls = new ArrayList<>();
  //			System.out.println(numTimesMaxChanged);
  		}
  	}
  	
  	static class ST{
  		static int[] theArray;
  		int l, r, min, minDex;
  		ST ll = null, rr = null;
  		
  		void print(boolean isRoot) {
  //			System.out.print("[at "+l+", "+r+"]");
  			if(l < r) {
  				ll.print(false);
  				rr.print(false);
  			}
  			else {
  				System.out.print(min+" ");
  			}
  			if(isRoot) System.out.println();
  		}
  		
  		ST(int l, int r){
  			this.l = l; this.r = r;
  			if(l < r) {
  				int m = l + (r - l) / 2;
  				ll = new ST(l, m);
  				rr = new ST(m+1, r);
  				refresh();
  			}
  			else {
  				min = theArray[l];
  				minDex = l;
  			}
  		}
  		
  		void refresh() {
  			if(l < r) {
  				if(ll.min < rr.min) {
  					min = ll.min;
  					minDex = ll.minDex;
  				}
  				else {
  					min = rr.min;
  					minDex =  rr.minDex;
  				}
  			}
  		}
  		
  		void update(int i, int newVal) {
  			if(i < l || r < i) return;
  //			System.out.println("[at "+l+", "+r+"]");
  			if(l < r) {
  				ll.update(i, newVal);
  				rr.update(i, newVal);
  				refresh();
  			}
  			else {
  				min = newVal;
  			}
  		}
  		
  		int queryMinDex(int l, int r) {
  //			System.out.println("[at "+this.l+", "+this.r+"]");
  			if(r < this.l || this.r < l) {
  //				System.out.println("out");
  				return -1;
  			}
  			if(l <= this.l && this.r <= r) {
  //				System.out.println("got it");
  				return minDex;
  			}
  			else {
  //				System.out.println("recurse");
  			}
  			
  			int ansL = ll.queryMinDex(l, r);
  			int ansR = rr.queryMinDex(l, r);
  
  //			System.out.println("[back at "+this.l+", "+this.r+"]");
  			int ans;
  			if(ansL == -1) ans = ansR;
  			else if(ansR == -1) ans = ansL;
  			else if(theArray[ansL] < theArray[ansR]) ans = ansL;
  			else ans = ansR;
  			
  //			System.out.println(ans+", that is to say value "+theArray[ans]);
  			return ans;
  		}
  	}
  }
   