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
for _ in range(int(input())):
    n = int(input())
    arr = list(map(lambda x : int(x) - 1, input().split()))
    nxt = [[-1] * 26 for _ in  range(n)]
    prv = [[-1] * 26 for _ in range(n)]
    fst = [-1] * 26
    lst = [-1] * 26
    initc = [0] * 26
    for i in range(n):
        initc[arr[i]] += 1
        if fst[arr[i]] == -1:
            fst[arr[i]] = i
        else:
            for j in range(i - 1, -1, -1):
                if arr[j] == arr[i]:
                    prv[i][arr[i]] = j
                    nxt[j][arr[i]] = i
                    break
    for i in range(n - 1, -1, -1):
        if lst[arr[i]] == -1:
            lst[arr[i]] = i
    ans = 0
    for i in range(26):
        cnt = initc[:]
        ans = max(cnt[i], ans)
        cnt[i] = 0
        if fst[i] == -1:
            continue
        l = fst[i]
        r = lst[i]
        if l == r:
            ans = max(1, ans)
            continue
        real_l = 0
        real_r = n - 1
        prsf = 2
        ans = max(ans, max(cnt))
        while l < r:
            while l > real_l:
                cnt[arr[real_l]] -= 1
                real_l += 1
            while r < real_r:
                cnt[arr[real_r]] -= 1
                real_r -= 1
            ans = max(ans, max(cnt) + prsf)
            l = nxt[l][i]
            r = prv[r][i]
            prsf += 2
    print(ans)
