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;
int countInRange(vector<int> &v, int L, int R) {
  auto it1 = lower_bound(v.begin(), v.end(), L);
  auto it2 = upper_bound(v.begin(), v.end(), R);
  if (it2 != v.begin()) {
    it2--;
    if (it1 != v.end() && *it1 >= L && *it1 <= R && *it2 >= L && *it2 <= R) {
      return it2 - it1 + 1;
    }
  }
  return 0;
}
void solve() {
  int n;
  cin >> n;
  int arr[n];
  vector<int> vec[250];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    vec[arr[i]].push_back(i);
  }
  int res = 0;
  for (int i = 1; i <= 27; i++) res = max(res, (int)vec[i].size());
  for (int i = 1; i <= 27; i++) {
    for (int len = 1; len <= vec[i].size(); len++) {
      int start = vec[i][len - 1], end = vec[i][vec[i].size() - len];
      for (int j = 1; j <= 27; j++) {
        if (i == j) continue;
        if (start < end) {
          int key = countInRange(vec[j], start, end);
          res = max(res, 2 * len + key);
        }
      }
    }
  }
  cout << res << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
