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
from collections import Counter, defaultdict

import sys

t = int(sys.stdin.readline())

for _ in xrange(t):
    n = int(sys.stdin.readline())
    arr = map(lambda x: int(x) - 1, sys.stdin.readline().split())
    n = len(arr)

    counts = [[0 for j in xrange(26)] for i in xrange(n + 1)]

    # pos = {}

    # prev_pos = {}
    for i in xrange(n):
        for j in xrange(26):
            counts[i + 1][j] = counts[i][j]
        counts[i + 1][arr[i]] += 1

        # if arr[i] in pos:
        #     prev_pos[i] = pos[arr[i]]
        #
        # pos[arr[i]] = i

    best = 1

    for i in xrange(n):
        for j in xrange(i + 1, n):
            if arr[i] == arr[j]:
                a = arr[i]
                l = i
                r = j
                left_a = counts[l][a] + 1
                right_a = counts[n][a] - counts[r][a]
                mid_b = max([counts[r][b] - counts[l][b] for b in xrange(26) if b != a])
                best = max(best, min(left_a, right_a) * 2 + mid_b)

    best = max(best, max(counts[n][i] for i in xrange(26)))

    print best





