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):
    n = int(input())

    a = list(map(int, input().split()))

    m = max(a)
    mem = [[0]*n for i in range((m+1))]
    position = {}
    #print(mem)

    for i in range(n):
        if a[i] not in position:
            position[a[i]] = []
        position[a[i]].append(i)
        for j in range(m+1):
            #print(j,i)
            mem[j][i] = mem[j][i-1]
        mem[a[i]][i] += 1

    ans = 0
    #print(mem)
    #print(position)
    for i in range(m+1):
        if i in position:
            ans = max(ans, len(position[i]))

            for p in range(len(position[i])):
                left = position[i][p] 
                right = position[i][len(position[i]) - p - 1] 
                #print(left,right)
                if left>right:
                    break
                max_middle = 0
                for middle in range(m+1):
                    if i == middle:
                        continue
                    freq = mem[middle][right] - mem[middle][left]

                    max_middle = max(max_middle, freq)
                if max_middle!=0:
                    ans = max(ans, 2*(p+1)+max_middle)
                
    print(ans)
