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
from bisect import bisect_left as lower_bound, bisect_right as upper_bound
from sys import stdin, stdout

def main():
    t = int(stdin.readline())

    for _ in range(t):
        n = int(stdin.readline())
        a = list(map(int, stdin.readline().strip().split()))
        
        pre, su = {i: [] for i in range(27)}, {i: [] for i in range(27)}

        for i, num in enumerate(a):
            su[num].append(i)

        ret = 1

        for i, num in enumerate(a):
            pre[num].append(i)
            su[num].pop(0)

            x = min(len(pre[num]), len(su[num]))

            if x == 0: continue

            lo, hi = pre[num][x-1], su[num][len(su[num])-x]

            y = 0

            for j in range(27):
                p = lower_bound(su[j], hi)
                q = upper_bound(su[j], lo)

                y = max(y, p-q)

            #print(i, num, lo, hi, x, y)

            ret = max(ret, 2*x+y) 

        print(ret)


if __name__ == "__main__": main()
