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>
#pragma GCC optimize("O3")
using namespace std;
const double EPS = 1e-9;
const int N = 2e5 + 9;
int n, a[N], sum[N][205];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    map<int, int> m;
    set<int> s;
    int mx = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      mx = max(mx, a[i]);
      s.insert(a[i]);
      m[a[i]]++;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= mx; j++) {
        sum[i][j] = 0;
      }
    }
    for (int i = 1; i <= n; i++) {
      sum[i][a[i]] = 1;
      for (int j = 1; j <= mx; j++) {
        sum[i][j] += sum[i - 1][j];
      }
    }
    int ans = 0;
    for (auto p : m) ans = max(ans, p.second);
    for (auto x : s) {
      int l = 0, r = n + 1, c = 0;
      while (l < r) {
        while (1) {
          l++;
          if (a[l] == x || l > n) break;
        }
        while (1) {
          r--;
          if (a[r] == x || r == 0) break;
        }
        if (l < r) {
          c += 2;
          for (int i = 1; i <= mx; i++) {
            if (i == x) continue;
            ans = max(ans, c + sum[r][i] - sum[l][i]);
          }
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
