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.*;
import java.util.StringTokenizer;
import java.util.*;

public class Solution {
    private static final long mod = 1000000007;

    public static void main(String args[]) throws Exception {

        FastScanner sc = new FastScanner();
        PrintWriter out=new PrintWriter(System.out);
        int t;
        t = sc.nextInt();

        while (t-- > 0) {
            solve(sc,out);
        }
        out.close();

    }

    public static void solve(FastScanner sc, PrintWriter out) {

        int n = sc.nextInt();
        int dp[][] = new int[201][n];
        int a[] = new int[n];
        for(int i = 0;i<n;i++)
            a[i] = sc.nextInt();

        for(int i = 0;i<n;i++){
            for(int j = 1;j<=200;j++)
                if(a[i] == j)
                    dp[j][i] = (i==0? 0:dp[j][i - 1]) + 1;

                else
                    dp[j][i] = i==0? 0:dp[j][i-1];
        }

        int ans = 0;
        for(int i = 0;i<n;i++){
            int j = a[i];
            int count = dp[j][i];
            int check = dp[j][n-1] - count + 1;
            if(check <= count) continue;
            int k = binarySearch(i+1,n-1,dp[j],check);
            //getMax between j and k exclusive
            int max = 0;
            for(int m = 1;m<=200;m++){
                max = Math.max(max,  dp[m][k-1] - dp[m][i]);
            }
            ans = Math.max(ans, max + count * 2);
        }

        for(int i = 1;i<=200;i++)
            ans = Math.max(dp[i][n-1], ans);

        out.println(ans);


    }

    public static int binarySearch(int l, int r, int[] a, int value){
        int ans = -1;
        while(l<=r){
            int mid = (l+r)/2;
            if(value > a[mid])
                l = mid + 1;
            else{
                ans = mid;
                r = mid - 1;
            } 
        }
        if(ans > 0 && a[ans-1] == value)
            ans = -1;
        return ans;
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        public String next() {
            while (!st.hasMoreElements())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

    }
}
