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
def data(): return sys.stdin.readline().strip()
out=sys.stdout.write
def mdata(): return map(int, data().split())
from collections import defaultdict as dd

for t in range(int(data())):
    n=int(data())
    A=list(mdata())
    s=set(A)
    m=1
    for i in s:
        str,end=0,n-1
        cnt=0
        while str<end:
            while str<end:
                if A[str]==i:
                    break
                str+=1
            while str<end:
                if A[end]==i:
                    break
                end-=1
            if A[str]==A[end]==i:
                d = dd(int)
                for j in range(str+1,end):
                    d[A[j]]+=1
                a=2
                if str==end:
                    a-=1
                if len(d.keys())!=0:
                    m=max(m,2*cnt+a+max(d.values()))
                else:
                    m=max(m,2*cnt+a)
            else:
                break
            cnt+=1
            str+=1
            end-=1
    print(m)