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
#include <bits/stdc++.h>
using namespace std;
map<int, vector<int> > pos;
vector<int> arr;
int rep;
void get_max(int a, int b) {
  for (long long i = 0; i < pos[a].size(); i++) {
    for (long long j = 0; j < pos[a].size(); j++) {
      if (i > j) continue;
      vector<int>::iterator it;
      vector<int>::iterator it2;
      int l, r;
      if (pos[b].size() > 0) {
        it = lower_bound(pos[b].begin(), pos[b].end(), pos[a][i]);
        l = (it - pos[b].begin());
      } else
        l = 0;
      if (pos[b].size() > 0) {
        it2 = upper_bound(pos[b].begin(), pos[b].end(), pos[a][j]);
        r = (pos[b].end() - it2);
      } else
        r = 0;
      rep = max(rep, min(r, l) * 2 + ((int)j - (int)i + 1));
    }
  }
}
void solve() {
  pos.clear();
  arr.clear();
  set<int> unique;
  set<int>::iterator it;
  set<int>::iterator it2;
  rep = 0;
  int n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    pos[temp].push_back(i);
    arr.push_back(temp);
    unique.insert(temp);
  }
  for (it = unique.begin(); it != unique.end(); it++) {
    for (it2 = unique.begin(); it2 != unique.end(); it2++) {
      if (*it == *it2)
        rep = max(rep, (int)pos[*it].size());
      else
        get_max(*it, *it2);
    }
  }
  cout << rep << endl;
}
int main() {
  int Q = 1;
  cin >> Q;
  for (long long q = 0; q < Q; q++) solve();
  return 0;
}
