The only difference between easy and hard versions is constraints.

You are given a sequence a consisting of n positive integers.

Let's define a three blocks palindrome as the sequence, consisting of at most two distinct elements (let these elements are a and b, a can be equal b) and is as follows: [\underbrace{a, a, ..., a}_{x}, \underbrace{b, b, ..., b}_{y}, \underbrace{a, a, ..., a}_{x}]. There x, y are integers greater than or equal to 0. For example, sequences [], [2], [1, 1], [1, 2, 1], [1, 2, 2, 1] and [1, 1, 2, 1, 1] are three block palindromes but [1, 2, 3, 2, 1], [1, 2, 1, 2, 1] and [1, 2] are not.

Your task is to choose the maximum by length subsequence of a that is a three blocks palindrome.

You have to answer t independent test cases.

Recall that the sequence t is a a subsequence of the sequence s if t can be derived from s by removing zero or more elements without changing the order of the remaining elements. For example, if s=[1, 2, 1, 3, 1, 2, 1], then possible subsequences are: [1, 1, 1, 1], [3] and [1, 2, 1, 3, 1, 2, 1], but not [3, 2, 3] and [1, 1, 1, 1, 2].

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2000) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2000) — the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 26), where a_i is the i-th element of a. Note that the maximum value of a_i can be up to 26.

It is guaranteed that the sum of n over all test cases does not exceed 2000 (∑ n ≤ 2000).

Output

For each test case, print the answer — the maximum possible length of some subsequence of a that is a three blocks palindrome.

Example

Input


6
8
1 1 2 2 3 2 1 1
3
1 3 3
4
1 10 10 1
1
26
2
2 1
3
1 1 1


Output


7
2
4
1
1
3
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;


import java.util.*;


public class template {

	InputStream is;
	PrintWriter out;
	String INPUT = "";
	int mod = 998244353;

	
	void solve() {
		int t = ni();
		while(t-- > 0) {
			int n = ni();
			int a[] = na(n);
			int f[][] = new int[201][n];
			f[a[0]][0]++;
			
			for(int i = 1 ; i<n ; i++) {
				for(int j = 0 ; j<201 ; j++) 
					f[j][i] = f[j][i-1];
				f[a[i]][i]++;
			}
			//display2D(f);display2D(b);
			int max = 0;
			
			for(int i = 1 ; i<=200 ; i++) {
				int left = 0 , right = n-1;
				
				int ct = 0;
				while(left < right) {
					while(left < n && a[left] != i) left++;
					while(right >= 0 && a[right] != i) right--;
					if(left >= right) break;
					
					ct+=2;
					int m = 0;
					for(int j = 1 ; j<201 ; j++) {
						m = Math.max(m,f[j][right-1]-f[j][left]);
					}
					max = Math.max(max, m+ct);
					left++;
					right--;
				}
			}
			if(max == 0) max = 1;
			out.println(max);
		}
	}
	void run() throws Exception
	{
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
		//tr(System.currentTimeMillis()-s+"ms");
	}

	public static void main(String[] args) throws Exception { new template().run(); }

	private void sort(int[] arr, int start, int end){
		if(start >= end) return;
		int mid = start + (end - start) / 2;
		sort(arr, start, mid);
		sort(arr, mid +1, end);
		merge(arr, start, mid, end);
	}
	private void merge(int[] arr, int start, int mid, int end){
		int i = start, j = mid + 1;
		int[] help = new int[end - start + 1];
		int c = 0;
		while(i <= mid && j <= end){
			help[c++] = arr[j] < arr[i] ? arr[j++] : arr[i++];
		}
		while(i <= mid){
			help[c++] = arr[i++];
		}
		while(j <= end){
			help[c++] = arr[j++];
		}
		for(int k = 0; k < help.length; k++){
			arr[start + k] = help[k];
		}
	}
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }

	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }

	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}

	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}

	private int[][] na(int n , int m)
	{
		int[][] a = new int[n][m];
		for(int i = 0;i < n;i++) 
			for(int j = 0 ; j<m ; j++) a[i][j] = ni();
		return a;
	}

	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	void display2D(int a[][]) {
		for(int i[] : a) {
			for(int j : i) {
				out.print(j+" ");
			}
			out.println();
		}
	}

	int[][] creategraph(int n , int m) {
		int g[][] = new int[n+1][];

		int from[] = new int[m];
		int to[] = new int[m];
		int ct[] = new int[n+1];

		for(int i = 0 ; i<m; i++) {
			from[i] = ni();
			to[i] = ni();
			ct[from[i]]++;
			ct[to[i]]++;
		}

		int parent[] = new int[n+1];

		for(int i = 0 ; i<n+1 ; i++) g[i] = new int[ct[i]];

		for(int i = 0 ; i<m ; i++) {
			g[from[i]][--ct[from[i]]] = to[i];
			g[to[i]][--ct[to[i]]] = from[i];
		}

		return g;
	}

	static long __gcd(long a, long b)
	{

		if(b == 0) 
		{
			return a;
		}
		else
		{
			return __gcd(b, a % b);
		}
	}

	// To compute x^y under modulo m
	static long power(long x, long y, long p)
	{

		// Initialize result
		long res = 1;

		// Update x if it is more than or
		// equal to p
		x = x % p;

		while (y > 0)
		{

			// If y is odd, multiply x
			// with result
			if (y % 2 == 1)
				res = (res * x) % p;

			// y must be even now
			y = y >> 1; // y = y/2
		x = (x * x) % p;
		}

		return res;
	}

	// Function to find modular 
	// inverse of a under modulo m
	// Assumption: m is prime
	static long modInverse(long a, int m)
	{
		if (__gcd(a, m) != 1) {
			//System.out.print("Inverse doesn't exist");
			return -1;
		}

		else {

			// If a and m are relatively prime, then
			// modulo inverse is a^(m-2) mode m
			//			System.out.println("Modular multiplicative inverse is "
			//											+power(a, m - 2, m));
			return power(a, m - 2, m);
		}
	}

	static long nCrModPFermat(int n, int r,
			int p , long fac[])
	{

		// Base case
		if (r == 0)
			return 1;

		// Fill factorial array so that we
		// can find all factorial of r, n
		// and n-r
		long t = (fac[n]* modInverse(fac[r], p))%p;

		return ( (t* modInverse(fac[n-r], p))% p);
	}


	private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}