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;
const int N = 2e5 + 6;
const int M = 1e9 + 7;
int A[N];
int mat[N][201];
int n;
void solve() {
  cin >> n;
  vector<vector<int> > vec(201);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 201; j++) mat[i][j] = 0;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    mat[i][A[i]]++;
    vec[A[i]].push_back(i);
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < 201; j++) {
      mat[i][j] += mat[i - 1][j];
    }
  }
  int ans = 0;
  for (int i = 0; i < 201; i++) {
    int l = vec[i].size();
    for (int j = 1; j < l / 2 + 1; j++) {
      int x = vec[i][j - 1];
      int y = vec[i][l - j];
      y--;
      int mx = 0;
      for (int k = 0; k < 201; k++) {
        mx = max(mx, mat[y][k] - mat[x][k]);
      }
      ans = max(ans, 2 * j + mx);
    }
  }
  if (n == 1 || ans == 0)
    ans = 1;
  else if (n == 2) {
    if (A[0] == A[1])
      ans = 2;
    else
      ans = 1;
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
}
