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 os
import sys
from atexit import register
from io import BytesIO
sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
sys.stdout = BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
input = lambda: sys.stdin.readline().rstrip('\r\n')
raw_input = lambda: sys.stdin.readline().rstrip('\r\n')

t = int(input())
for _ in range(t):
    n = int(input())
    arr = map(int,raw_input().split(" "))
    maxv=  max(arr)
    ans = arr.count(maxv)
    dp =   [[0]*(1+n) for i in range(maxv+1)]
    dic = [ [] for i in range(maxv+1)]
    for i,v in enumerate(arr):
        dic[v].append(i)

    for b in range(1,maxv+1)[::-1]:
        for j in range(n):
            if arr[j] == b:
                dp[b][j+1] = dp[b][j]+ 1
            else:
                dp[b][j+1] = dp[b][j]
    for a in range(1,maxv+1):
        v = len(dic[a])
        for j in range(v/2):
            x,y = dic[a][j]+1,dic[a][v-1-j]+1
            vv = 0
            for i in range(1,maxv+1):
                vv = max(vv,dp[i][y-1]-dp[i][x])

            ans = max(ans,vv+(j+1)*2)
    print ans

