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
for _ in range(int(input())):
    n=int(input())
    a=list(map(int,input().split()))
    c=[[0 for i in range(n)]]
    for i in range(1,27):
        k=[]
        if a[0]==i:
            k.append(1)
            count=1
        else:
            k.append(0)
            count=0
        for j in range(1,n):
            if a[j]==i:
                count+=1
            k.append(count)
        c.append(k[:])
    maxx=0
    for i in range(1,27):
        maxxi=0
        for j in range(1,27):
            count=c[j][n-1]
            if count==0 or c[i][n-1]==0:
                continue
            else:
                for pi in range(0,n):
                    if a[pi]==i:
                        break
                for pj in range(n-1,-1,-1):
                    if a[pj]==i:
                        break
                nu=2
                while (pj-pi)>1:
                    count=max(count,nu+c[j][pj-1]-c[j][pi])
                    for o in range(pi+1,pj):
                        if a[o]==i:
                            break
                    pi=o
                    for o in range(pj-1,pi,-1):
                        if a[o]==i:
                            break
                    pj=o
                    nu+=2
            maxxi=max(maxxi,count)
        maxx=max(maxx,maxxi)
    print(maxx)