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
import math

if __name__ == '__main__':
    T = int(sys.stdin.readline())

    for _ in range(T):

        n = int(sys.stdin.readline())
        a = list(map(int, sys.stdin.readline().split()))

        pos = [[] for _ in range(27)]
        for i, x in enumerate(a):
            pos[x].append(i + 1)
        # for i, x in enumerate(pos):
        #     if x:
        #         print(i, ':', x)

        pre_sum = [[0] * (n + 1) if pos[i] else [] for i in range(27)]

        for i, x in enumerate(a):
            pre_sum[x][i + 1] = 1

        for x in range(1, 27):
            if pre_sum[x]:
                for i in range(1, n + 1):
                    pre_sum[x][i] += pre_sum[x][i - 1]
                # print(x, 'pre sum', pre_sum[x])

        ans = 0
        for x in range(1, 27):
            if pos[x]:
                x_num = len(pos[x])
                ans = max(ans, x_num)
                mid = x_num // 2
                for j in range(mid):
                    left = j
                    right = x_num - 1 - j
                    left_pos = pos[x][left]
                    # print('debug', x_num, mid, j, left, right)
                    right_pos = pos[x][right]
                    for y in range(1, 27):
                        if pos[y] and y != x:
                            ans = max(ans, pre_sum[y][right_pos] - pre_sum[y][left_pos] + (j + 1) * 2)
        print(ans)
