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;
using namespace std::chrono;
void pr_init() {}
long long dp[2005][27];
vector<vector<long long> > gp;
void solve() {
  long long t;
  cin >> t;
  for (long long te = 0; te < t; te++) {
    long long n;
    cin >> n;
    long long a[n + 1];
    gp.assign(27, vector<long long>());
    for (long long j = 1; j < n + 1; j++) {
      cin >> a[j];
      gp[a[j]].push_back(j);
    }
    for (long long i = 1; i < n + 1; i++) {
      for (long long j = 0; j < 27; j++) dp[i][j] = dp[i - 1][j];
      dp[i][a[i]]++;
    }
    long long len = 0;
    for (long long i = 1; i < 27; i++)
      if ((long long)gp[i].size() > 0) {
        for (long long k = 1; k <= 26; k++) {
          for (long long l = 0; l < (long long)gp[i].size(); l++) {
            for (long long z = 0; z <= l; z++) {
              len = max(len, l - z + 1 +
                                 2 * min(dp[gp[i][z] - 1][k],
                                         dp[n][k] - dp[gp[i][l]][k]));
            }
          }
        }
      }
    cout << len << endl;
  }
}
int32_t main() {
  pr_init();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  auto start = high_resolution_clock::now();
  solve();
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop - start);
  return 0;
}
