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;
void write(vector<int> &v) {
  for (auto i : v) cout << i << " ";
  cout << "\n";
}
void read(vector<int> &v) {
  for (auto &i : v) cin >> i;
}
const int INF = 1e9;
const int64_t INFF = 1e18;
const int N = 1e6 + 69;
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  read(v);
  vector<vector<int>> total_cnt(202);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    total_cnt[v[i]].push_back(i);
    ans = max((int)(total_cnt[v[i]]).size(), ans);
  }
  vector<int> run_cnt(202);
  for (int i = 0; i < n; i++) {
    int cnt = ++run_cnt[v[i]];
    int total = total_cnt[v[i]].size();
    if (total - 2 * cnt < 0) continue;
    int last = total_cnt[v[i]][total - cnt];
    for (int j = 1; j < 201; j++) {
      if (j == v[i]) continue;
      int f = lower_bound((total_cnt[j]).begin(), (total_cnt[j]).end(), i + 1) -
              total_cnt[j].begin();
      int l = lower_bound((total_cnt[j]).begin(), (total_cnt[j]).end(), last) -
              total_cnt[j].begin();
      ans = max(ans, 2 * cnt + l - f);
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
