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
import collections
import threading
import string

def sparsetable(a):
    tmp = [0]*201
    st = []
    for i in range(len(a)):
        tmp[a[i]] += 1
        st.append(tmp[:])
    return st

def func(x, y, st):
    if x>=y: return 0
    
    ans = 0
    for i in range(201):
        ans = max(ans, st[y][i] - st[x][i] )
    return ans

def main():

    testn = int(input())
    for _ in range(testn):
        n = int( input() )
        a = list(map(int, input().split() ))
        d = collections.defaultdict(list)
        st = sparsetable(a)
        for i in range(n):
            d[ a[i] ].append(i)
        ans = 1
        for key in d:
            z = len(d[key])
            l = d[key]
            for i in range(z//2):
                ans = max(ans, (i+1)*2 + func(l[i], l[z-1-i]-1, st))

        print(ans)
        
        


input = sys.stdin.readline
main()