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();
signed main() {
  clock_t beg = clock();
  cout << fixed << setprecision(15);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  int tt = t;
  while (t--) {
    solve();
  }
  cerr << "\nExecution time: " << (clock() - beg) / 1000 << '\n';
  return 0;
}
const int N = 2e5 + 5;
const int A = 200;
int n;
int a[N];
vector<int> pos[205];
int pref[N][205];
int get(int x, int y) { return pref[y][x]; }
void solve() {
  cin >> n;
  int mx = 0;
  for (int a = (1); a <= (200); ++a) pref[0][a] = 0;
  for (int i = (1); i <= (n); ++i) {
    cin >> a[i];
    for (int b = (1); b <= (200); ++b) pref[i][b] = pref[i - 1][b];
    ++pref[i][a[i]];
    pos[a[i]].push_back(i);
    mx = max(mx, (int)pos[a[i]].size());
  }
  for (int a = (1); a <= (200); ++a) {
    reverse((pos[a]).begin(), (pos[a]).end());
  }
  for (int i = (1); i <= (n); ++i) {
    int cnt = get(a[i], i);
    if (cnt <= get(a[i], n) - get(a[i], i)) {
      int l = i + 1;
      int r = pos[a[i]][cnt - 1] - 1;
      int here = 0;
      for (int b = (1); b <= (200); ++b) {
        here = max(here, get(b, r) - get(b, l - 1));
      }
      mx = max(mx, here + cnt + cnt);
    }
  }
  cout << mx << '\n';
  for (int i = (1); i <= (200); ++i) pos[i].clear();
}
