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
t=int(input())

for _ in range(t):
    res=0
    n=int(input())
    a=[int(i)-1 for i in input().split()]
    ll,lr=[[0]*201],[[0]*201]
    ll[0][a[0]]=1
    lr[0][a[-1]]=1
    lookup=[[-1]*(n+1) for i in range(201)]
    c=[0]*201
    for i in a:
        c[i]+=1
    for i in a[1:]:
        ll.append(ll[-1][:])
        ll[-1][i]+=1
    
    for i in a[:-1][::-1]:
        lr.append(lr[-1][:])
        lr[-1][i]+=1
    lr=lr[::-1]
    
    temp=[0]*201
    for i in range(n-1,-1,-1):
        temp[a[i]]+=1
        lookup[a[i]][temp[a[i]]]=i
    res=max(temp)
    for i in range(201):
        for j in range(n):
            v=ll[j][i]
            j2=lookup[i][v]
            if j2<=j:continue
            tres=0

            for k in range(201):
                t=c[k]-ll[j][k]-lr[j2][k]
                if tres<t:
                    tres=t
            tres+=2*v
            res=max(res,tres)
    print(res)




    