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 maxx = 2e3 + 7;
int cas, n;
int a[maxx];
vector<int> pos[maxx];
int main() {
  cin >> cas;
  while (cas--) {
    for (int i = 1; i <= 26; i++) pos[i].clear();
    int ans = 0;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      pos[a[i]].push_back(i);
    }
    for (int i = 1; i <= 26; i++) {
      int res = 0;
      int sz = pos[i].size();
      if (sz <= 1) continue;
      int l = 0, r = sz - 1;
      while (pos[i][l] < pos[i][r]) {
        res += 2;
        int dd = 0;
        for (int j = 1; j <= 26; j++) {
          int cnt = 0;
          for (int k = 0; k < pos[j].size(); k++) {
            if (pos[j][k] > pos[i][l] && pos[j][k] < pos[i][r]) cnt++;
          }
          dd = max(dd, cnt);
        }
        ans = max(ans, res + dd);
        l++, r--;
      }
    }
    if (!ans) ans = 1;
    cout << ans << endl;
  }
  return 0;
}
