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
def compute(a,b,isAdd = True):
    ans = [0] * len(a)
    for i in range(0,len(a)):
        if isAdd:
            ans[i] = a[i] + b[i]
        else :
            ans[i] = a[i] - b[i]
    return ans

def solve(arr):
    n = len(arr)
    if(n==0):
        print(0)
        return
    distinctElements = list(set(arr))
    d = len(distinctElements)
    mapElementToIndex = {}
    for i in range(0,d):
        mapElementToIndex[distinctElements[i]] = i
    totalFre = [0] * d

    for i in range(0,len(arr)):
        totalFre[mapElementToIndex[arr[i]]]+=1

    presentCount = [0] * d
    rigthIndex = [n] * d
    rightFre  = [[0] * d for i in range(0,d)]

    leftFre = [0] * d

    finalAns = 0
    for i in range(0,n):
        index = mapElementToIndex[arr[i]]
        leftFre[index] +=1
        r = rigthIndex[index]
        for j in range(r-1,i,-1):
            if arr[j] == arr[i]:
                rigthIndex[index] = j
                presentCount[index] += 1
                rightFre[index][index]+=1
                #TODO compute ans
                inbetweenFre = compute(totalFre,compute(leftFre,rightFre[index]),False)
                ans = 2*presentCount[index] + max(inbetweenFre)
                finalAns = max(ans,finalAns)
                break   
            else:
                rightFre[index][mapElementToIndex[arr[j]]]+=1

    print(max(1,finalAns))

t= int(input())
for i in range(0,t):
    input()
    arr = list(map(lambda x:int(x),input().split()))
    solve(arr)