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
# import bisect
# import os
# import io
# from collections import Counter
import bisect
from collections import defaultdict
# import math
# import random
# import heapq as hq
# from math import sqrt
import sys
from functools import reduce, cmp_to_key
from collections import deque
import threading
from itertools import combinations


# sys.setrecursionlimit(200000)
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline


def input():
    return sys.stdin.readline().strip()


def iinput():
    return int(input())


def tinput():
    return input().split()


def rinput():
    return map(int, tinput())


def rlinput():
    return list(rinput())


# mod = int(1e9)+7


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))


# ----------------------------------------------------
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

if __name__ == "__main__":
    for _ in range(iinput()):
        n = iinput()
        a = rlinput()
        d = defaultdict(list)
        for i in range(n):
            d[a[i]].append(i)
        ans = 1
        for i in d:
            temp = d[i]
            l = len(temp)
            ans = max(ans, len(temp))
            # print(i)
            for j in range(l//2):
                start = temp[j]
                end = temp[l - j - 1]
                # print(i,start,end)
                for k in d:

                    if k != i:
                        tt = d[k]
                        # if tt[-1]<start:
                        #     continue
                        # print(start,end,'fg')
                        val = bisect.bisect(tt, end) - \
                            bisect.bisect(tt, start)
                        ans = max(ans, val+2*(j+1))
            # print('=========================')
        print(ans)
        # print('---------------------------------------')
