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
def solve(arr,n,ans):
    dp = [[0 for i in range(27)] for j in range(n)]
    for i in range(n):
        dp[i][arr[i]] += 1
        if i > 0:
            for j in range(27):
                dp[i][j] += dp[i-1][j]

    max_val = 0
    for i in range(n):
        left = dp[i][:]
        for j in range(i+1,n):
            total = 0
            counts = dp[j-1][:]
            for k in range(27):
                counts[k] -= dp[i][k]

            total += max(counts)
            right = dp[-1][:]
            for k in range(27):
                right[k] -= dp[j-1][k]

            max_curr = 0
            for k in range(27):
                max_curr = max(max_curr,2*min(left[k],right[k]))
                
            total += max_curr

            max_val = max(max_val,total)

        max_val = max(max_val,max(left))

    ans.append(max_val)

def main():
    t = int(input())
    ans = []
    for i in range(t):
        n = int(input())
        arr = list(map(int,input().split()))
        solve(arr,n,ans)

    for i in ans:
        print(i)


main()
