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 distribution(n):
    return (n-1)//2

def problem_A():
    print(distribution(int(input())))

s = 'abcdefghijklmnopqrstuvwxyz'
def cons_str(n,b):
    a = s[:b]
    ans = ''
    while n > b:
        ans += a
        n -= b
    ans += a[:n]
    return ans

def problem_B():
    n,a,b = list(map(int,input().split()))
    print(cons_str(n,b))

def comp_teams(a):
    if len(a) < 2:
        return 0
    s = len(set(a))
    if s == len(a):
        return 1
    a.sort()
    t = 0
    temp = 1
    for i in range(1,len(a)):
        if a[i] == a[i-1]:
            temp += 1
        else:
            if t < temp:
                t = temp
            temp = 1
    if t < temp:
        t = temp
    
    if s <= t - 1:
        return s
    if s == t:
        return s - 1
    else:
        return t

def problem_C():
    l = int(input())
    a = list(map(int,input().split()))
    print(comp_teams(a))

def antisudoku(a):
    for i in range(9):
        for j in range(9):
            if a[i][j] == '2':
                a[i][j] = '1'
    return a

def problem_D():
    a = []
    for _ in range(9):
        a.append(list(input()))
    b = antisudoku(a)
    for i in range(9):
        print(''.join(b[i]))

def three_block_palindrome(s):
    a = []
    b = []
    c = []
    l = len(s)
    for _ in range(l):
        a = []
        for _ in range(l+1):
            a.append([0]*26)
    l += 1
    for i in range(1,l):
        k = s[i-1] - 1 
        for j in range(26):
            a[i][j] = a[i-1][j]
        a[i][k] += 1
    maxlength = 0
    for i in range(l):
        for j in range(i,l):
            maxa = 0
            maxb = 0
            for k in range(26):
                if a[l-1][k]-a[j][k] > a[i][k]:
                    if maxa < a[i][k]:
                        maxa = a[i][k]
                else:
                    if maxa < a[l-1][k]-a[j][k]:
                        maxa = a[l-1][k]-a[j][k]
                    
                if maxb < a[j][k] - a[i][k]:
                        maxb = a[j][k] - a[i][k]
            if maxlength < 2*maxa + maxb:
                maxlength = 2*maxa + maxb

    return maxlength
                
def problem_E1():
    l = int(input())
    s = list(map(int,input().split()))
    print(three_block_palindrome(s))

cases = int(input())
for _ in range(cases):
    # problem_A()
    # problem_B()
    # problem_C()
    # problem_D()
    problem_E1()