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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n, 0);
    vector<vector<int>> totsofar(200, vector<int>(n, 0));
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      a[i] = x - 1;
      for (int j = 0; j < 200; j++) {
        if (i != 0) totsofar[j][i] = totsofar[j][i - 1];
        if (a[i] == j) totsofar[j][i]++;
      }
    }
    vector<map<int, int>> indback(200);
    for (int i = n - 1; i >= 0; i--) {
      int x = a[i];
      indback[x].insert({totsofar[x][n - 1] - totsofar[x][i] + 1, i});
    }
    int maxm = 0;
    for (int i = 0; i < 200; i++) {
      if (maxm < totsofar[i][n - 1]) maxm = totsofar[i][n - 1];
    }
    for (int i = 0; i < n; i++) {
      int x = a[i];
      if (totsofar[x][i] > (totsofar[x][n - 1] / 2)) continue;
      int y = indback[x][totsofar[x][i]];
      for (int j = 0; j < 200; j++) {
        int z = totsofar[j][y - 1] - totsofar[j][i];
        int tot = (z + (2 * totsofar[x][i]));
        if (maxm < tot) maxm = tot;
      }
    }
    cout << maxm << '\n';
  }
  return 0;
}
