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 io
import os
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from sys import stdout
if __name__=="__main__":
    for t in range(int(input())):
        n=int(input())
        a=list(map(int,input().strip().split()))
        b={}
        c={}
        mal = 0
        for i in range(n):
            if b.get(a[i])==None:
                b[a[i]]=[i]
                c[a[i]]=[]
            else:
                b[a[i]].append(i)
        for i in c:
            for j in range(n):
                if j!=0:
                    c[i].append(c[i][-1]+(a[j]==int(i)))
                else:
                    c[i].append(int((a[j]==int(i))))
            mal =max(mal,c[i][-1])
        if mal!=n:
            for k in b:
                i=0
                j=len(b[k])-1
                l =b[k][i]
                r =b[k][j]
                while l<r and i<j:
                    ma=0
                    for m in b:
                        if m!=k:
                            x=c[int(m)][l]
                            y=c[int(m)][r]
                            ma=max(ma,y-x)
                    mal=max(2*(i+1)+ma,mal)
                    i = i + 1
                    j = j - 1
                    l=b[k][i]
                    r=b[k][j]
        stdout.write(str(mal)+"\n")