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
// No sorceries shall prevail. //           
import java.util.*;
import java.io.*;
public class InVoker {
	//Variables
	static long mod = 1000000007;
	static long mod2 = 998244353;
	static FastReader inp= new FastReader();
	static PrintWriter out= new PrintWriter(System.out);
	public static void main(String args[]) {		    	
	   	InVoker g=new InVoker();
	   	g.main();
	   	out.close();
	}
	
	//Main
	void main() {
		
		int t=inp.nextInt();
		while(t-->0) {
			int n=inp.nextInt();
			int a[]=new int[n];
			input(a,n);
			int gg=1;
			int freq[][]=new int[n+1][26];
			for(int i=0;i<n;i++) {
				for(int j=0;j<26;j++) {
					freq[i+1][j]=freq[i][j];
				}
				freq[i+1][a[i]-1]++;
				gg=Math.max(gg, freq[i+1][a[i]-1]);
			}
			//out.println(gg+"=gg");
			for(int i=0;i<n-2;i++) {
				for(int j=i+1;j<n-1;j++) {
					int max=0;
					for(int k=0;k<26;k++) {
						if(freq[i+1][k]>0 && freq[n][k]-freq[j+1][k]>0) {
							for(int p=0;p<26;p++) {
								if(p!=k && (freq[j+1][p]-freq[i+1][p])>0) {
									max=Math.max( (freq[j+1][p]-freq[i+1][p]) + 2*Math.min((freq[n][k]-freq[j+1][k]),freq[i+1][k]),max);
								}		
							}
						}	
					}
					gg=Math.max(max, gg);
				}
			}
			out.println(gg);
		}
		
	}
	
	
	
	
	
	
	
	
	
	
	
	
	
	// Classes
	static class Edge implements Comparable<Edge>{
		int l,r;
		Edge(){}
		Edge(int l,int r){
			this.l=l;
			this.r=r;
		}
		@Override
		public int compareTo(Edge e) {
			return (l-e.l)!=0?l-e.l:r-e.r;
		}
	}
	
    static class Segment implements Comparable<Segment> {
		long l, r, initialIndex;
		Segment () {}
		Segment (long l_, long r_, long d_) {
		    this.l = l_;
		    this.r = r_;
		    this.initialIndex = d_;
		}
		@Override
		public int compareTo(Segment o) {
		    return (int)((l - o.l) !=0 ? l-o.l : initialIndex - o.initialIndex);
		}
    }
    
    static class FastReader { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
        String next() { 
            while (st==null || !st.hasMoreElements()) { 
                try { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
        int nextInt() { 
            return Integer.parseInt(next()); 
        } 
        long nextLong() { 
            return Long.parseLong(next()); 
        } 
        double nextDouble() { 
            return Double.parseDouble(next()); 
        } 
        String nextLine() { 
            String s=""; 
            try { 
                s=br.readLine(); 
            } 
            catch (IOException e) { 
                e.printStackTrace(); 
            } 
            return s; 
        } 
    } 
    
    // Functions
	static long gcd(long a, long b) { 
		return b==0?a:gcd(b,a%b);
	}
	static int gcd(int a, int b) { 
		return b==0?a:gcd(b,a%b);
	}
	
	void reverse(long[] A,int l,int r) {
        int i=l,j=r-1;
        while(i<j) {
            long t=A[i];
            A[i]=A[j];
            A[j]=t;
            i++;j--;
        }
    }
    void reverse(int[] A,int l,int r) {
        int i=l,j=r-1;
        while(i<j) {
            int t=A[i];
            A[i]=A[j];
            A[j]=t;
            i++;j--;
        }
    }
    
    //Input Arrays
    static void input(long a[], int n) {
		for(int i=0;i<n;i++) {
			a[i]=inp.nextLong();
		}
	}
	static void input(int a[], int n) {
		for(int i=0;i<n;i++) {
			a[i]=inp.nextInt();
		}
	}	
	static void input(String s[],int n) {
		for(int i=0;i<n;i++) {
			s[i]=inp.next();
		}
	}
	static void input(int a[][], int n, int m) {
		for(int i=0;i<n;i++) {
			for(int j=0;j<m;j++) {
				a[i][j]=inp.nextInt();
			}
		}
	}
	static void input(long a[][], int n, int m) {
		for(int i=0;i<n;i++) {
			for(int j=0;j<m;j++) {
				a[i][j]=inp.nextLong();
			}
		}
	}
		
}
