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 pii = std::pair<int, int>;
using pll = std::pair<long long, long long>;
const int maxk = 203;
std::vector<int> prefix[maxk];
std::vector<int> at[maxk];
void solve() {
  int n;
  std::cin >> n;
  for (int i = 0; i < maxk; i++) {
    prefix[i].resize(n + 1);
    prefix[i][0] = 0;
    at[i].clear();
  }
  for (int i = 1; i <= n; i++) {
    int x;
    std::cin >> x;
    for (int j = 0; j < maxk; j++) {
      prefix[j][i] = prefix[j][i - 1];
    }
    prefix[x][i]++;
    at[x].push_back(i);
  }
  int ans = 0;
  for (int a = 1; a < maxk; a++) {
    ans = std::max(ans, (int)at[a].size());
    int l = 0;
    int r = at[a].size() - 1;
    while (l < r) {
      int lt = at[a][l];
      int rt = at[a][r];
      ++l, --r;
      int need = l;
      ans = std::max(ans, 2 * need);
      for (int b = 1; b < maxk; b++) {
        int has = prefix[b][rt - 1] - prefix[b][lt];
        ans = std::max(ans, 2 * need + has);
      }
    }
  }
  std::cout << ans << "\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  int t;
  std::cin >> t;
  while (t--) solve();
  return 0;
}
