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;
void solve() {
  int n;
  cin >> n;
  int len = 26;
  int arr[n][len];
  memset(arr, 0, sizeof(arr));
  vector<int> addres[len];
  int x;
  cin >> x;
  x--;
  addres[x].push_back(0);
  arr[0][x]++;
  for (int i = 1; i < n; ++i) {
    cin >> x;
    x--;
    addres[x].push_back(i);
    for (int ele = 0; ele < len; ++ele) arr[i][ele] = arr[i - 1][ele];
    arr[i][x]++;
  }
  int mx_len = 0;
  for (int i = 0; i < len; ++i) mx_len = max(mx_len, arr[n - 1][i]);
  for (int x = 0; x < len; ++x) {
    int l = 0, r = addres[x].size() - 1;
    while (l < r) {
      int li = addres[x][l], ri = addres[x][r];
      int x_ele = (l + 1) * 2;
      int m_ele = 0;
      for (int ty = 0; ty < len; ++ty) {
        int tmp_y = arr[ri - 1][ty] - arr[li][ty];
        m_ele = max(m_ele, tmp_y);
      }
      mx_len = max(mx_len, x_ele + m_ele);
      l++;
      r--;
    }
  }
  cout << mx_len << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}
