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;
class Solution {
 public:
  void solve(int num, std::istream& in, std::ostream& out) {
    int n;
    in >> n;
    vector<int> arr(n);
    vector<vector<int>> pz(200);
    for (int i = 0; i < n; i++) {
      in >> arr[i];
      arr[i]--;
      pz[arr[i]].push_back(i);
    }
    int mx = 0;
    for (auto& vt : pz) mx = max(mx, (int)vt.size());
    for (int i = 0; i < 200; i++) {
      for (int j = 0; j < 200; j++) {
        if (i == j || pz[i].size() < 2 || pz[j].empty()) continue;
        int lo = 0, hi = pz[i].size() - 1;
        while (lo < hi) {
          int st = pz[i][lo], en = pz[i][hi];
          auto it1 = upper_bound(pz[j].begin(), pz[j].end(), en);
          auto it0 = upper_bound(pz[j].begin(), pz[j].end(), st);
          int left = 0;
          if (it0 == pz[j].end()) {
            left = pz[j].size();
          } else if (it0 != pz[j].begin()) {
            it0--;
            left += 1 + it0 - pz[j].begin();
          }
          int right = 0;
          if (it1 == pz[j].end()) {
            right = pz[j].size();
          } else if (it1 != pz[j].begin()) {
            it1--;
            right += 1 + it1 - pz[j].begin();
          }
          int mid = right - left;
          mx = max(mx, 2 * (lo + 1) + mid);
          lo++, hi--;
        }
      }
    }
    out << mx << endl;
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  int T;
  in >> T;
  for (int i = 0; i < T; i++) solution.solve(i + 1, in, out);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}
