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() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int i, j, n, ans = 0, k1, k2;
    cin >> n;
    long long int arr[n];
    vector<vector<long long int> > vec(26);
    for (i = 0; i < n; i++) {
      cin >> arr[i];
      vec[arr[i] - 1].push_back(i);
    }
    long long int dp1[n][26], dp2[n][26];
    for (i = 0; i < n; i++)
      for (j = 0; j < 26; j++) {
        dp1[i][j] = 0;
        dp2[i][j] = 0;
      }
    dp1[0][arr[0] - 1]++;
    for (i = 1; i < n; i++) {
      for (j = 0; j < 26; j++)
        if (arr[i] == j + 1)
          dp1[i][j] += dp1[i - 1][j] + 1;
        else
          dp1[i][j] += dp1[i - 1][j];
    }
    dp2[n - 1][arr[n - 1] - 1]++;
    for (i = n - 2; i >= 0; i--) {
      for (j = 0; j < 26; j++)
        if (arr[i] == j + 1)
          dp2[i][j] += dp2[i + 1][j] + 1;
        else
          dp2[i][j] += dp2[i + 1][j];
    }
    for (i = 0; i < 26; i++) {
      long long int dd = vec[i].size();
      ans = max(ans, dd);
    }
    for (i = 0; i < 26; i++) {
      for (k1 = 0; k1 < vec[i].size(); k1++) {
        for (k2 = k1; k2 < vec[i].size(); k2++) {
          long long int d1 = vec[i][k1], d2 = vec[i][k2];
          for (j = 0; j < 26; j++)
            if (i != j)
              ans = max(ans, k2 - k1 + 1 + 2 * min(dp1[d1][j], dp2[d2][j]));
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
