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
range = xrange
input = raw_input

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0

t = inp[ii]; ii += 1
out = []
for _ in range(t):
    n = inp[ii]; ii += 1
    A = [a - 1 for a in inp[ii: ii + n]]; ii += n
    cur = [0]*200
    B = [list(cur)]
    C = [[] for _ in range(200)]
    for i in range(n):
        a = A[i]
        cur[a] += 1
        B.append(list(cur))
        C[a].append(i)

    ans = max(len(occ) for occ in C)
    for a in range(200):
        occ = C[a]
        for j in range(len(occ)//2):
            D1 = B[occ[j] + 1]
            D2 = B[occ[~j]]
            for k in range(200):
                ans = max(ans, 2 * (j + 1) + D2[k] - D1[k])
    out.append(ans)
print '\n'.join(str(x) for x in out)
