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 math
import sys

input = sys.stdin.buffer.readline

def solveAll():
    nbCases = int(input())
    for _ in range(nbCases):
        case = readCase()
        print(solve(case))

def readCase():
    input()
    return [int(x) for x in input().split()]

def solve(nums):
    indices = [[] for _ in range(201)]
    for i, num in enumerate(nums):
        indices[num] += [i]
    
    bucket = [0] * 201
    for num in nums:
        bucket[num] += 1

    ans = -math.inf
    for num in range(1, 201):
        ans = max(ans, computeThreeBlock(nums, indices[num], bucket.copy()))
    return ans

def computeThreeBlock(nums, indices, bucket):
    le = 0
    ri = len(nums) - 1

    indicesFrom = 0
    indicesTo = len(indices) - 1
    ans = -math.inf
    while indicesFrom <= indicesTo:
        while le <= indices[indicesFrom]:
            bucket[nums[le]] -= 1
            le += 1
        while ri >= indices[indicesTo]:
            bucket[nums[ri]] -= 1
            ri -= 1
        if indicesFrom < indicesTo:
            ans = max(ans, max(bucket) + 2* (indicesFrom + 1))
        else:
            ans = max(ans, max(bucket) -1 + 2* (indicesFrom + 1))
        indicesFrom += 1
        indicesTo -= 1
    return ans

solveAll()
