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 i in range(t):
    n = int(input())
    a = [int(i) for i in input().split()]
    new_a = []
    now = a[0]
    s = 1
    for i in range(1, n):
        if a[i] == now:
            s += 1
        else:
            new_a.append([now, s])
            s = 1
            now = a[i]
    new_a.append([now, s])
    p = {}
    a = list(set(a))
    for i in range(len(a)):
        p[a[i]] = [0]
    for i in range(len(new_a)):
        for j in p.keys():
            if j == new_a[i][0]:
                p[j].append(p[j][-1] + new_a[i][1])
            else:
                p[j].append(p[j][-1])
    ans = -1
    for i in range(len(new_a)):
        for j in range(i, len(new_a)):
            mx = -1
            good_color = -1
            for k in a:
                counter_color = p[k][j + 1] - p[k][i]
                if counter_color > mx:
                    mx = max(mx, counter_color)
                    good_color = k
            for k in a:
                first = p[k][i] - p[k][0]
                second = p[k][len(new_a)] - p[k][j + 1]
                ans = max(ans, mx + 2 * min(first, second))
    for i in a:
        ans = max(ans, p[i][-1])
    print(ans)