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
var t = parseInt(readline());

function div(val, by){
    return (val - val % by) / by;
}

function setCharAt(str,index,chr) {
    if(index > str.length-1) return str;
    return str.substr(0,index) + chr + str.substr(index+1);
}

var ch = 'a';
var acode = ch.charCodeAt(0);


for (var tt = 0; tt < t; tt++) {
    var n = parseInt(readline());
    var a = readline().split(' ').map(Number);
    var m = {};
    for (var i = 0; i < n; i++) {
        m[a[i]] = m[a[i]] ? m[a[i]] + 1 : 1;
    }
    var nums = Object.keys(m).map(Number);
    var res = 1;
    for (var i = 0; i < nums.length; i++) {
        var num = nums[i];
        var l = 0;
        var r = n - 1;
        var mcopy = Object.assign({}, m);
        var locallen = 0;
        while (l < r) {
            while (l < n && a[l] != num) {
                mcopy[a[l]] -= 1;
                l += 1;
            }
            while (r >= 0 && a[r] != num) {
                mcopy[a[r]] -= 1;
                r -= 1;
            }
            if (l >= r) {
                break;
            }
            mcopy[a[l]] -= 1;
            l += 1;
            mcopy[a[r]] -= 1;
            r -= 1;
            locallen += 2;
            var localmax = 0;
            var mcopykeys = Object.keys(mcopy);
            for (var j = 0; j < mcopykeys.length; j++) {
                var localkey = mcopykeys[j];
                if (mcopy[localkey] > localmax) {
                    localmax = mcopy[localkey];
                }
            }
            
            if (locallen + localmax > res) {
                res = locallen + localmax;
            }
        }
    }
    
    
    write(res+'\n');
}
