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 n;
vector<int> a;
int f() {
  cin >> n;
  vector<int> g[200];
  vector<int> s[200];
  for (int i = 0; i < 200; ++i) s[i].assign(n + 1, 0);
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
    g[a[i]].push_back(i);
    s[a[i]][i + 1] = 1;
  }
  for (int i = 0; i < 200; ++i)
    for (int j = 1; j < n + 1; ++j) s[i][j] += s[i][j - 1];
  int res = 0;
  for (int i = 0; i < 200; ++i) res = max(res, s[i].back());
  for (int e = 0; e < 200; ++e) {
    vector<int>& z = g[e];
    if (z.size() < 2) continue;
    int l = 0;
    int r = z.size() - 1;
    while (l < r) {
      int add = 0;
      int fr = z[l] + 1;
      int to = z[r] - 1;
      for (int i = 0; i < 200; ++i) add = max(add, s[i][to + 1] - s[i][fr]);
      res = max(res, 2 * (l + 1) + add);
      ++l;
      --r;
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) cout << f() << "\n";
  return 0;
}
