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
from array import array  # noqa: F401
import typing as Tp  # noqa: F401


def input():
    return sys.stdin.buffer.readline().decode('utf-8')


def main():
    t = int(input())
    ans = [0] * t
    acc = [array('i', [0]) * (2 * 10**5 + 1) for _ in range(200)]

    for ti in range(t):
        n = int(input())
        a = (-1,) + tuple(map(lambda x: int(x) - 1, input().split())) + (-1, )
        keys = sorted(set(a[1:-1]))

        for i, v in enumerate(a[1:-1], start=1):
            for j in keys:
                acc[j][i] = acc[j][i - 1]
            acc[v][i] += 1

        maxv = max(acc[i][n] for i in keys)

        for x in keys:
            l, r, outside = 0, n + 1, 1
            while l < r:
                while l < r and a[l] != x:
                    l += 1
                while l < r and a[r] != x:
                    r -= 1
                if l == r:
                    break

                inside = 0
                for k in keys:
                    inside = max(inside, acc[k][r - 1] - acc[k][l])
                maxv = max(maxv, outside * 2 + inside)

                l, r, outside = l + 1, r - 1, outside + 1

        ans[ti] = maxv

    sys.stdout.buffer.write(('\n'.join(map(str, ans)) + '\n').encode('utf-8'))


if __name__ == '__main__':
    main()
