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 io,os
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
from collections import defaultdict
def solve(n,A):
    if n==1:
        return 1
    if n==2:
        if len(set(A))==1:
            return 2
        return 1
    L=[[0]*27 for _ in range(n+1)]
    for i,a in enumerate(A):
        for j in range(27):
            L[i+1][j]=L[i][j]
        L[i+1][a]+=1
    ans=0
    for l in range(n):
        for r in range(l,n):
            Left,Right=[0]*27,[0]*27
            mid=0
            for i in range(27):
                Left[i]=L[l][i]
                mid=max(mid,L[r+1][i]-L[l][i])
                Right[i]=L[n][i]-L[r+1][i]
            tmp=0
            for i in range(27):
                tmp=max(tmp,min(Left[i],Right[i]))
            #print(l,r,tmp,mid)
            ans=max(ans,mid+tmp*2)
    return ans

def main():
    t=int(input())
    for _ in range(t):
        n=int(input())
        A=tuple(map(int,input().split()))
        ans=solve(n,A)
        sys.stdout.write(str(ans)+'\n')
    
if __name__=='__main__':
    main()