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
'''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
'''

import sys
import math
import copy
from collections import Counter

debug = 1
readln = sys.stdin.readline
#sys.setrecursionlimit(1000000)

def readint():
    return int(readln().strip())

def readints():
    return map(int, readln().split())

def readstr():
    return readln().strip()

def readstrs():
    return readln().split()

def dbg(*args):
    if debug: print(' '.join(map(str, args)))

def subr(i, j, cter):
    if j < i:
        return 0
    elif j == i:
        return 1
    else:
        c = cter[j] - (cter[i-1] if i > 0 else Counter())
        return c.most_common(1)[0][1]


def solve(n,A):
    maxa = 0
    pos = [[] for _ in range(201)]
    cter = []
    for i, a in enumerate(A):
        pos[a].append(i)
        maxa = max(maxa, a)
        if i > 0:
            cter.append(cter[i-1].copy())
        else:
            cter.append(Counter())
        cter[i][a] += 1

    keys = [a for a in range(len(pos)) if pos[a]]
    ans = 0
    for x in keys:
        posx = pos[x]
        lenx = len(posx)
        for i in range(len(posx) // 2):
            ans = max(ans, (i+1)*2 + subr(posx[i]+1, posx[lenx-i-1]-1, cter))

    return max(ans, subr(0, n-1, cter))

t = readint()
for _ in range(t):
    n = readint()
    A = readints()
    print(solve(n,A))