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 + 10;
long long a[N] = {};
map<int, int> nums;
void io() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
int main(int argc, char const *argv[]) {
  io();
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    nums.clear();
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      nums[a[i]]++;
    }
    long long res = 1;
    for (int i = 1; i <= 26; i++) {
      int l = 0, r = n - 1, cnt = 0;
      int vis[30] = {};
      res = max(res, 1ll * nums[i]);
      while (l < r) {
        while (a[l] != i && l < r) {
          vis[a[l]]++;
          l++;
        }
        if (l == r) break;
        while (a[r] != i && l < r) {
          vis[a[r]]++;
          r--;
        }
        if (l == r) break;
        cnt += 2;
        l++;
        r--;
        for (int j = 1; j <= 26; j++) {
          if (i == j) continue;
          res = max(res, 1ll * cnt + nums[j] - vis[j]);
        }
      }
    }
    cout << res << endl;
  }
  return 0;
}
