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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, t, k, maxk, l, bk, bl, ans, aa;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<vector<int> > a(200, vector<int>(0));
    for (int i = 0; i < n; ++i) {
      cin >> aa;
      a[aa - 1].push_back(i);
    }
    ans = 0;
    for (int i = 0; i < 200; ++i) {
      for (int j = 0; j < 200; ++j) {
        if (i == j) continue;
        if (a[i].size() == 0 || a[j].size() == 0) {
          ans =
              ans < a[i].size() + a[j].size() ? a[i].size() + a[j].size() : ans;
          continue;
        }
        k = 0;
        l = a[i].size() - 1;
        while (k < l) {
          bk = 0;
          bl = a[j].size() - 1;
          while (bk <= bl) {
            if (a[i][k] > a[j][bk]) {
              bk++;
              continue;
            }
            if (a[i][l] < a[j][bl]) {
              bl--;
              continue;
            }
            break;
          }
          if (bk > bl) break;
          ans = max(ans, 2 * (k + 1) + bl - bk + 1);
          k++;
          l--;
        }
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
