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
for f in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    d = {}
    for i in a:
        d[i] = d.get(i, 0) + 1
    ans = 0
    d1 = {}
    for i in range(n):
        if d1.get(a[i], 0) + 1 <= d[a[i]] // 2:
            maxx = 0
            d2 = {}
            d1[a[i]] = d1.get(a[i], 0) + 1
            x = d1[a[i]]
            j = i + 1
            while d[a[i]] - d2.get(a[i], 0) - d1[a[i]] >= x:
                if a[i] == a[j]:
                    d2[a[i]] = d2.get(a[i], 0) + 1
                else:
                    d2[a[j]] = d2.get(a[j], 0) + 1
                    maxx = max(maxx, d2[a[j]])
                j += 1
            ans = max(ans, maxx + 2 * x)
        else:
            ans = max(ans, d[a[i]])
    print(ans)

