import java.util.*;
import java.io.*;

import static java.lang.Math.*;

public class Practice {

	static Scanner scn = new Scanner(System.in);
	static StringBuilder sb = new StringBuilder();

	public static void main(String[] HastaLaVistaLa) {
		int t = scn.nextInt();
		// int t = 1;
		for(int tests = 0; tests < t; tests++) solve();
		System.out.println(sb);
	}

	// main function is written here
	public static void solve() {
		int n = scn.nextInt();
		int[] a = new int[n];
		for(int i = 0; i < n; i++) a[i] = scn.nextInt();
		int i = 0, j = n - 1;
		while(i <= j && a[i] == a[j]) {
			i++; j--;								// trimming all a(i) == a(n - 1 - i) from both sides of course
		}
		if(j <= i || pal(a, a[i], i, j) || pal(a, a[j], i, j)) { // if either is true, then YES otherwise NO
			sb.append("YES");
		}else {
			sb.append("NO");
		}
		sb.append("\n");
	}

	// This function checks if the array is palindrome after removing the remove element i.e. a[i] or a[j] (for which a[i] != a[j])
	// Approach: I decided to make a new list of integer a[i, j] excluding remove element, then checked if it's palindrome or not
	public static boolean pal(int[] a, int remove, int i, int j) {
		List<Integer> l = new ArrayList<>();
		for(int k = i; k <= j; k++) if(a[k] != remove) l.add(a[k]);
		int left = 0, right = l.size() - 1;
		while(left < right){
			if(l.get(left) != l.get(right)) 
				return false;					 // found a(i) != a(n - 1 - i)
			left++; right--;
		} 
		return true;
	}
}