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;
inline void R_P(vector<int> &vec) {
  int temp;
  cin >> temp;
  vec.push_back(temp);
}
inline bool checkMax(int &base, int test) {
  return (test > base) ? base = test, true : false;
}
inline bool checkMin(int &base, int test) {
  return (test < base) ? base = test, true : false;
}
template <class T>
void test(vector<T> &vec, const string &warn) {
  cout << warn << '\n';
  for (int i = (0); i < (vec.size()); i++) cout << vec[i] << ' ';
  cout << "\nTest end..............\n";
}
vector<int> vec;
int mid;
int bor;
vector<int> pre;
vector<int> tmp;
void Copy(vector<int> &cop) {
  cop.clear();
  for (int X : vec) {
    if (X == mid || X == bor) cop.emplace_back(X);
  }
}
void Prefixing() {
  pre.clear();
  pre.emplace_back(0);
  for (int X : tmp) {
    if (X == mid)
      pre.emplace_back(pre[pre.size() - 1] + 1);
    else
      pre.emplace_back(pre[pre.size() - 1]);
  }
}
int BF() {
  if (tmp.empty()) return 0;
  int l = -1, r = (int)tmp.size();
  int xl = 0, xr = 0;
  int len = 0;
  int ans = 0;
  while (l < r) {
    while (xl < len && l < (int)tmp.size() - 1) {
      l++;
      if (tmp[l] == bor) xl++;
    }
    while (xr < len && r > 0) {
      r--;
      if (tmp[r] == bor) xr++;
    }
    if (l < r && xl == len && xr == len) {
      ans = max(ans, 2 * len + pre[r] - pre[l + 1]);
    }
    len++;
  }
  return ans;
}
void solve() {
  int n;
  cin >> n;
  vec.clear();
  for (int i = (0); i < (n); i++) {
    int tmp;
    cin >> tmp;
    vec.emplace_back(tmp);
  }
  int ans = 0;
  for (int i = (1); i <= 26; i++)
    for (int j = (i + 1); j <= 26; j++) {
      mid = i, bor = j;
      Copy(tmp);
      Prefixing();
      ans = max(ans, BF());
      mid = j, bor = i;
      Copy(tmp);
      Prefixing();
      ans = max(ans, BF());
    }
  cout << ans << '\n';
}
int main() {
  std::ios::sync_with_stdio(false), cin.tie(NULL);
  int t;
  cin >> t;
  for (int i = (0); i < (t); i++) solve();
  return 0;
}
