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
# python3

#===============================================================================
from collections import defaultdict, Counter
from functools import lru_cache
from heapq import heappush, heappop
from math import gcd, floor, ceil
from sys import stdin, stdout

def ilist():
    return [int(x) for x in stdin.readline().strip().split(" ")]
def iint():
    return int(stdin.readline().strip())
def istr():
    return stdin.readline().strip()
#===============================================================================

def diff(l2, l1):
    d = [0] * (26 + 1)
    for i in range(len(l1)):
        d[i] = l2[i] - l1[i]
    return d

def solve(a):
    n = len(a)
    count = [[0] * (26 + 1) for _ in range(n + 1)]
    for i in range(n):
        count[i+1] = count[i][:]
        count[i+1][a[i]] += 1
    #for e in count:
    #    print(e[:4])
    res = max(count[n])
    for i in range(1, n+1):
        for j in range(i+1, n):
            l = diff(count[i], count[0])
            m = diff(count[j], count[i])
            r = diff(count[n], count[j])
            #print(l[:4], m[:4], r[:4], i, j, n)
            outer = 0
            inner = max(m)
            for j in range(1, 26 + 1):
                if l[j] == r[j]:
                    outer = max(outer, 2*l[j])
            res = max(res, outer + inner)
    return res


if __name__ == '__main__':
    t = iint()
    for _ in range(t):
        n = iint()
        a = ilist()
        print(solve(a))
