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, stdout
T = int(stdin.readline())
res = ''


for p in xrange(T):
    ans = ''
    n = input()
    arr = map(int, raw_input().split())
    dp = [[0]*27 for i in xrange(27)]
    psums = [[0]*27 for i in xrange(n)]

    # for i in psums:
    #     print i

    # print "\n"

    for i in xrange(n):
        for j in xrange(1, 27):
            psums[i][j] = psums[i-1][j]
        psums[i][arr[i]] += 1

    # for i in psums:
    #     print i

    # print "\n"

    m = 0
    for i in xrange(1, 27):
        for j in xrange(1, 27):
            
            l = 0
            r = n-1
            while l <= r and l < n and r >= 0:
                if l<n and r>=0 and arr[l] == i and arr[r] == i:
                    # print l, r, i, j
                    a = b = c = 0
                    if i != j:
                        if psums[r][j]-psums[l][j] != 0:
                            a = psums[r][j]-psums[l][j]+1
                        if psums[l][i] < psums[r][i]:
                            b = psums[l][i]+psums[n-1][i]-psums[r][i]
                        m = max(m, a+b)
                    else:
                        m = max(m, psums[n-1][i])

                    l += 1
                    r -= 1
                if not(l <= r and l < n and r >= 0):
                    break
                if arr[l] != i:
                    l += 1
                if arr[r] != i:
                    r -= 1

    res+=str(m)+'\n'

stdout.write(res)
