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 sys import stdin
###############################################################
def iinput(): return int(stdin.readline())
def minput(): return map(int, stdin.readline().split())
def linput(): return list(map(int, stdin.readline().split()))
###############################################################

from collections import defaultdict
from bisect import bisect

t = iinput()
while t:
    t-=1
    n = iinput()
    a = linput()
    d = defaultdict(list)
    for i in range(n):
        d[a[i]].append(i)
    ans = 1
    for p in d:
        ans = max(ans, len(d[p]))
        for q in d:
            if p!=q:
                for i in range(len(d[p])//2):
                    x = bisect(d[q], d[p][i])
                    y = bisect(d[q], d[p][len(d[p])-1-i])
                    # for j in range(len(d[q])):
                    #     if d[q][j] < d[p][len(d[p])-1-i]:
                    #         y = j
                    # for j in range(len(d[q])-1, -1, -1):
                    #     if d[q][j] > d[p][i]:
                    #         x = j
                    if y!=x:
                        ans = max(ans, y-x+2*(i+1))
    print(ans)

