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(0);
  cout.tie(0);
  int t, a[2000], p[2001][26];
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < 26; ++i) p[0][i] = 0;
    for (int i = 0; i < n; ++i) {
      for (int k = 0; k < 26; ++k) p[i + 1][k] = p[i][k];
      cin >> a[i];
      ++p[i + 1][a[i] - 1];
    }
    int ans = 0;
    for (int k = 1; k <= 26; ++k) {
      int low = 0, high = n - 1, m = 0;
      while (low < high) {
        while (low < n && a[low] != k) ++low;
        while (high >= low && a[high] != k) --high;
        if (low < high) {
          m += 2;
          ++low;
          --high;
          for (int i = 0; i < 26; ++i)
            ans = max(ans, m + p[high + 1][i] - p[low][i]);
        }
      }
    }
    cout << max(ans, 1) << '\n';
  }
  return 0;
}
