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
input = raw_input

from collections import Counter
def answer():
    n = int(input())
    nums = map(int, input().split(" "))
    
    c = Counter(nums)
    nums_set = c.keys()

    out = max(c.values()) # single

    # double
    set_len = len(nums_set)
    for i in range(set_len):
        x1 = nums_set[i]
        x1_kosuu = c[x1]
        for j in range(i+1, set_len):
            x2 = nums_set[j]
            x2_kosuu = c[x2]

            cur_nums = [num for num in nums if num == x1 or num == x2]
            
            x1_counts = [None] * x1_kosuu
            x2_counts = [None] * x2_kosuu

            x1_count = 0
            x2_count = 0
            for x in cur_nums:
                if x == x1: 
                    x1_counts[x1_count] = x2_count
                    x1_count += 1
                else:
                    x2_counts[x2_count] = x1_count
                    x2_count += 1
            
            i = 0
            j = x1_kosuu-1
            while i < j:
                tmp1 = i+1 + x1_kosuu-j
                tmp2 = x1_counts[j] - x1_counts[i]
                tmp_out = tmp1 + tmp2
                out = max(out, tmp_out)
                i += 1
                j -= 1

            i = 0
            j = x2_kosuu-1
            while i < j:
                tmp1 = x2_counts[j] - x2_counts[i]
                tmp2 = i+1 + x2_kosuu-j
                tmp_out = tmp1 + tmp2
                out = max(out, tmp_out)
                i += 1
                j -= 1


    return out

t = int(input())
for _ in range(t):
    print(answer())