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 collections import Counter

def solve(new):
    ans=0
    x=0
    cnt=Counter(new)
    y = cnt[1]
    i=0
    j=len(new)-1
    while (i<j):
        
        # ans=max(ans,2*x+y)
        # print(i,j,x,y,ans)
        
        if new[i]==0:
            if new[j]==0:
                x+=1
                i+=1
                j-=1
            else:
                y-=1
                j-=1
        else:
            y-=1
            i+=1
        ans=max(ans,2*x+y)
        # print(i,j,x,y,ans)
    return ans
        
    
def ri():
    return int(input())
 
def rl():
    return list(map(int, input().split()))
    
t = ri()
for _ in range(t):
    n =ri()
    numbers= rl()
    cnt = Counter(numbers)


    ans=0
    for a in cnt.keys():


        for b in cnt.keys():
            if a==b:
                ans=max(cnt[a],ans)
            else:
                new = []
                for i in numbers:
                    if i==a:
                        new.append(0)
                    if i==b:
                        new.append(1)
                ans = max(ans,solve(new))
    print(ans)

                
                
                