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

sys.setrecursionlimit(50000)
input = sys.stdin.readline


def input_list():
    return map(int, input().split())

def r(n_i, n_j, i, j, m=[]):
    #print(m)
    n = n_i + n_j
    if n_i == 0:
        return n_j
    elif n_j == 0:
        return n_i
    else:
        if m[0] == m[n-1] and m[0] == i:
            return 2 + r(n_i-2, n_j, i, j, m[1:n-1])
        elif m[0] == j:
            return max(n_j, r(n_i, n_j-1, i, j, m[1:n]))
        else:
            return max(n_j, r(n_i, n_j-1, i, j, m[0:n-1]))

def f():
    n = int(input())
    l = list(input_list())
    f_result = 0
    s = set(l)
    for i in s:
        for j in s:
            #print(i, j)
            m = []
            n_i = 0
            n_j = 0
            for k in l:
                #print(k)
                if k == i:
                    m.append(k)
                    n_i += 1
                elif k == j:
                    m.append(k)
                    n_j += 1
            f_result = max(f_result, r(n_i, n_j, i, j, m))
    print(f_result)



def main():
    t = int(input())
    for _ in range(t):
        f()


main()
