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 collections import deque

def solve2_d(a):
    d1 = deque(i for i in range(len(a)) if a[i] == 1)
    d2 = deque(i for i in range(len(a)) if a[i] == 2)
    
    d1_left = []
    d1_right = []
    best = len(d2)
    
    while len(d1) >= 2:
        d1_left.append(d1.popleft())
        d1_right.append(d1.pop())
        while len(d2) > 0 and d2[0] < d1_left[-1]:
            d2.popleft()
        while len(d2) > 0 and d2[-1] > d1_right[-1]:
            d2.pop()
        best = max(best, len(d1_left) + len(d1_right) + len(d2))
    
    return best

t = int(input())
for i in range(t):
    n = int(input())
    a = [int(v) for v in input().split()]
    # ~ print("a: ", a)
    best = -1
    for x in range(1, 26 + 1):
        for y in range(1, 26 + 1):
            # ~ print(x, y)
            a2 = [v for v in a if v == x or v == y]
            a2 = [(1 if v == x else 2) for v in a2]
            best = max(best, sum(1 for v in a2 if v == 1))
            best = max(best, sum(1 for v in a2 if v == 2))
            if x != y:
                best = max(best, solve2_d(a2))
    print(best)

# ~ print(solve2_d([1,2,1,2,1]))
