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 sys,bisect
from collections import defaultdict
input = sys.stdin.readline

############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
    return(int(input().rstrip()))
def inlt():
    return(list(map(int,input().rstrip().split())))
def insr():
    s = input().rstrip()
    return(s[:len(s) - 1])
def invr():
    return(map(int,input().rstrip().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############

t = inp()
for _ in range(t):
    n = inp()
    nums = inlt()

    if n == 1:
        print(1)
    elif n == 2:
        if len(set(nums)) == 1:
            print(2)
        else:
            print(1)
    else:
        ans = 1
        for i in range(1,27):
            #print(i)
            left = 0
            right = n - 1
            cnt = 0
            while left < right:
                mid = defaultdict(int)
                if nums[left] == i:
                    while left < right:
                        if nums[right] == i:
                            cnt += 2
                            for j in range(left + 1,right):
                                mid[nums[j]] += 1
                            cur_max = 0
                            for key in mid:
                                if mid[key] > cur_max:
                                    cur_max = mid[key]
                            #print('left:{} right: {} mid{}'.format(left,right,mid))
                            ans = max(ans,cnt + cur_max)
                            right -= 1
                            break
                        else:
                            right -= 1
                left += 1 
        print(ans)
