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
from collections import deque, defaultdict, Counter
from bisect import bisect_left, bisect #bisect(list, num, beg, end)
import math
def solve(inf):
    n = inf[0]
    a = list(map(int, input().split()))
    uni = list(set(a))
    if len(uni) == 1:
        return n
    ans = 0
    count = defaultdict(lambda:0)
    indxs = defaultdict(lambda:[])
    for i in range(len(a)):
        indxs[a[i]].append(i)
    pre = []
    for i in range(len(a)):
        count[a[i]] += 1
        pre.append(count.copy())
    pre.append(defaultdict(lambda:0))
    # pre.append(pre[-1])
    # print(newa)
    # pre.append([0,0])
    # print(pre)
    for key in indxs:
        l = 0
        r = len(indxs[key]) - 1
        ans = max(ans, r + 1)
        for i in range((r+1)//2):
            p = indxs[key][i]
            q = indxs[key][r - i]
            # if key == a: aaa...bbbb...aaa
            # print((p, q))
            for newkey in indxs:
                if newkey == key:
                    continue
                # print('key', key, 'newkey', newkey, pre[q-1][newkey] - pre[p][newkey], 2*(i+1))
                ans = max(ans, pre[q-1][newkey] - pre[p][newkey] + 2*(i+1))
    return ans
for _ in range(int(input())):
    inf = list(map(int, input().split()))

    print(solve(inf))
