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())
maxa=27

for _ in range(t):
    n=int(input())
    ar=[int(x) for x in input().split()]
    tbc=[]
    maxleng=0
    leng=1

    ftot=[[0 for j in range(n)] for i in range(maxa+1)]
    btot=[[0 for j in range(n)] for i in range(maxa+1)]

    for i in range(n):
        for j in range(1,maxa+1):
            ftot[j][i]=ftot[j][i-1]
            if ar[i]==j:
                ftot[j][i]+=1

    arrev=ar[::-1]
    for i in range(n):
        for j in range(1,maxa+1):
            btot[j][i]=btot[j][i-1]
            if arrev[i]==j:
                btot[j][i]+=1

    for i in range(len(btot)):
        btot[i]=btot[i][::-1]

    for i in range(n):
        for j in range(i,n):
            if ar[i]==ar[j]:
                val=ar[i]
                leng=1+ftot[val][j]-ftot[val][i]
                maxext=0
                for k in range(1,maxa+1):
                    if k!=val:
                        ext=2*min(ftot[k][i],btot[k][j])
                        if ext>maxext:
                            maxext=ext
                leng+=maxext
                if leng>maxleng:
                    maxleng=leng


    print(maxleng)

