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 pre[200001][201];
int a[200001];
int n;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  for (T = 1; T <= 200; T++) {
    pre[0][T] = 0;
  }
  cin >> T;
  while (T--) {
    int i, j, k, l, m, r, mid;
    cin >> n;
    vector<vector<int>> gr(201);
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      gr[a[i]].emplace_back(i);
    }
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= 200; j++) {
        pre[i][j] = pre[i - 1][j];
      }
      pre[i][a[i]]++;
    }
    int ans = 0;
    for (k = 1; k <= 200; k++) {
      int sz = gr[k].size();
      ans = max(ans, sz);
      i = 0;
      j = sz - 1;
      while (i < j) {
        mid = 0;
        l = gr[k][i];
        r = gr[k][j];
        for (m = 1; m <= 200; m++) {
          if (m != k) mid = max(mid, pre[r][m] - pre[l - 1][m]);
        }
        ans = max(ans, i + 1 + sz - j + mid);
        i++;
        j--;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
