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;
const int maxn = 1e6 + 10;
vector<int> a[maxn];
void solve() {
  int n, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    a[x].push_back(i);
  }
  long long res = 0;
  for (int i = 1; i <= 200; i++) {
    if (!a[i].empty()) {
      long long len = a[i].size();
      res = max(res, len);
      for (long long l = 0; l < len / 2; l++) {
        long long r = len - 1 - l;
        long long b = a[i][l], c = a[i][r];
        for (int j = 1; j <= 200; j++) {
          if (!a[j].empty() && i != j) {
            res = max(res,
                      (l + 1) * 2 + (lower_bound(a[j].begin(), a[j].end(), c) -
                                     lower_bound(a[j].begin(), a[j].end(), b)));
          }
        }
      }
    }
  }
  for (int i = 1; i <= 200; i++) a[i].clear();
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
    if (t) cout << endl;
  }
  return 0;
}
