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 r(a, b):
    n = len(a)
    s = len(a) // 2
    j = s - 1
    v = -1
    while (j >= 0):
        x = a[j]
        y = a[n - 1 - j]
        c=b[y]-b[x+1]
        s = (2 * (j + 1) + c)
        if s > v:
            v = s
        j += -1
    return v


t = int(input())
for i in range(t):
    n = int(input())
    b = list(map(int, input().split()))
    a = []
    c = dict()
    for j in range(1, 201):
        c[j] = 0
    for j in range(202):
        a.append([])
    for j in range(n):
        a[b[j]].append(j)
        c[b[j]] = 1
    pre=[[0] for i in range(201)]
    for i in range(1,201):
        if c[i]==1:
            y=a[i]
            d=dict()
            for j in y:
                d[j]=1
            j=0
            while(j<n):
                if j in d.keys():
                    pre[i].append(pre[i][-1]+1)
                else:
                    pre[i].append(pre[i][-1])
                j+=1



    m = -1
    d = []
    for j in range(1, 201):
        if c[j] == 1:
            d.append(j)
    for j in d:
        for p in d:
            if p == j:
                h = len(a[p])
            else:
                h = r(a[j], pre[p])
            if h > m:
                m = h

    print(m)