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
def leftBinarySearch(arr,l,r,x):
    temp=-1
    while l<=r:
        mid=(l+r)//2
        if arr[mid]<x:
            temp=mid
            l=mid+1
        else:
            r=mid-1
    return temp
def rightBinarySearch(arr,l,r,x):
    temp=-1
    while l<=r:
        mid=(l+r)//2
        if arr[mid]>x:
            temp=mid
            r=mid-1
        else:
            l=mid+1
    return temp
t=int(input())
for _ in range(t):
    n=int(input())
    arr=list(map(int,input().split()))
    dict1={}
    ind={}
    for i in range(n):
        if arr[i] not in dict1.keys():
            dict1[arr[i]]=1
            ind[arr[i]]=[i]
        else:
            dict1[arr[i]]+=1
            ind[arr[i]].append(i)
    ans=0
    for i in dict1.keys():
        ans=max(ans,dict1[i])
    for i in dict1.keys():
        a=dict1[i]
        for j in ind.keys():
            if j!=i:
                x=0
                y=dict1[i]-1
                for k in range(a//2):
                    ind1=rightBinarySearch(ind[j],0,dict1[j]-1,ind[i][x])
                    ind2=leftBinarySearch(ind[j],0,dict1[j]-1,ind[i][y])
                    if ind1!=-1 and ind2!=-1 and ind2>=ind1:
                        ans=max(ans,(ind2-ind1+1)+(2*(x+1)))
                    x+=1
                    y-=1
    print(ans)