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
t = int(input())
for _ in range(t):
    n = int(input())
    D = {}
    DC = {}
    DS = {}
    A = list(map(int, input().split()))
    for i in range(n):
        if A[i] in D:
            D[A[i]].append(i)
            DS[A[i]][i] = 1
            DC[A[i]] += 1
        else:
            D[A[i]] = [i]
            DS[A[i]] = [0 for __ in range(n)]
            DS[A[i]][i] = 1
            DC[A[i]] = 1
    for i in DS:
        for k in range(n-1):
            DS[i][k+1] += DS[i][k]
    M = max(DC[i] for i in DC)
    for i in D:
        m = DC[i]
        for j in D:
            if i != j:
                for k in range(1, m//2+1):
                    l, r = D[i][k-1], D[i][-k]
                    if l != 0:
                        cnt = DS[j][r-1] - DS[j][l-1]
                    else:
                        cnt = DS[j][r-1]
                    if M < cnt + k * 2:
                        M = cnt + k * 2
    print(M)