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
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)

def getCnts(j,l,r):
    if l>r:
        return 0
    if l==0:
        return prefCnt[r][j]
    else:
        return prefCnt[r][j]-prefCnt[l-1][j]

t=int(input())
for _ in range(t):
    n=int(input())
    a=[int(x) for x in input().split()]
    
    prefCnt=[[0 for _a in range(27)] for _b in range(n)]
    for j in range(1,27):
        cnts=0
        for i in range(n):
            if a[i]==j:
                cnts+=1
            prefCnt[i][j]=cnts
#    print(prefCnt)
    ans=0
    for outer in range(1,27):
        outerLen=0
        l=-1
        r=n
        while l<r:
            for inner in range(1,27):
                innerLen=getCnts(inner,l+1,r-1)
                ans=max(ans,outerLen*2+innerLen)
#                print('outer:{} inner:{} l:{} r:{} totalLength:{}'.format(outer,inner,l,r,outerLen*2+innerLen))
            l+=1
            while l<n and a[l]!=outer:
                l+=1
            r-=1
            while r>-1 and a[r]!=outer:
                r-=1
            outerLen+=1
#    print('here')
    print(ans)