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;
vector<vector<int>> v;
int lbd(int idx, int no) {
  return lower_bound(v[idx].begin(), v[idx].end(), no) - v[idx].begin() - 1;
}
int ubd(int idx, int no) {
  return upper_bound(v[idx].begin(), v[idx].end(), no) - v[idx].begin();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t, n, tmp;
  cin >> t;
  while (t--) {
    int ans = 1;
    v = vector<vector<int>>(27);
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      cin >> tmp;
      v[tmp].push_back(i);
    }
    for (int i = 1; i <= 26; ++i) {
      if (v[i].size() <= 1) continue;
      int s = 0, e = v[i].size() - 1, cnt = 1;
      while (s < e) {
        for (int j = 1; j <= 26; ++j)
          ans = max(ans, lbd(j, v[i][e]) - ubd(j, v[i][s]) + 1 + 2 * cnt);
        ++s;
        --e;
        ++cnt;
      }
    }
    cout << ans << '\n';
  }
}
