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
numbers = 26
for _ in range(int(input())):
    n = int(input())
    seq = [int(x)-1 for x in input().split()]  #TODO
    pre = [(n+1)*[0] for i in range(numbers)]
    for j, val in enumerate(seq, 1):
        for i in range(numbers):
            if val == i:
                pre[i][j] = pre[i][j-1] + 1
            else:
                pre[i][j] = pre[i][j-1]
    ans = 1
    for i in range(numbers):  # for x value
        max_val = pre[i][-1] // 2
        for x in range(1, 1+max_val):  # for min x len
            bmin = pre[i].index(x)  # leftmost
            bmax = n+1-pre[i][::-1].index(pre[i][-1]-x)  # right most 
            # print(pre[i],x,bmin, bmax)
            ymax = 0
            if bmax - bmin > 1:       
                for j in range(numbers):  # for y
                    ymax = max(ymax, pre[j][bmax-1] - pre[j][bmin])
            # print('sol', i,x,ymax)
            ans = max(ans, ymax + 2*x)
    # print("ANS ", end='')
    print(ans) 
            