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;
[[maybe_unused]] const int MOD = 1000000007;
[[maybe_unused]] const int INTMAX = numeric_limits<int>::max();
void writeln() { cout << "\n"; }
template <class T>
inline void print(T&& a);
template <class T>
inline void priws(T&& a);
template <class T>
inline void read(T& a);
template <class... Args>
inline void readln(Args&... args) {
  (read(args), ...);
}
template <class H, class... T>
inline void writeln(H&& h, T&&... t) {
  priws(h);
  (print(t), ...);
  writeln();
}
void run() {
  int m = 202;
  int n;
  readln(n);
  vector<int> a(n);
  readln(a);
  vector<vector<int>> c(n + 1, vector<int>(m, 0));
  vector<vector<int>> d(n + 1, vector<int>(m, 0));
  vector<vector<int>> counts(m + 1, vector<int>(n + 1, n));
  for (int i = 0; i < (int)(n); ++i) c[i + 1] = c[i], c[i + 1][a[i]]++;
  for (int i = n - 1; i >= (int)(0); --i)
    d[i] = d[i + 1], counts[a[i]][++d[i][a[i]]] = i;
  int ans = 0;
  for (int left = 0; left < (int)(n); ++left)
    for (int index = max(0, left - 2); index < (int)(min(left + 2, n)); ++index)
      if (int j = a[index], x = c[left][j], right = counts[j][x];
          right > left) {
        int temp = 0;
        for (int k = 1; k < (int)(m); ++k)
          temp = max(temp, c[right][k] - c[left][k]);
        ans = max(ans, x * 2 + temp);
      }
  writeln(ans);
}
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  readln(t);
  for (int i = 0; i < (int)(t); ++i) run();
  cerr << fixed << setprecision(0)
       << "Execution time = " << 1000.0 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
template <class T1, class T2>
inline ostream& operator<<(ostream& os, pair<T1, T2> const& _a);
template <
    typename T, typename D = decltype(*begin(declval<T>())),
    typename enable_if<!is_same<T, basic_string<char>>::value>::type* = nullptr>
ostream& operator<<(ostream& os, T const& _a) {
  auto it = begin(_a);
  if (it != end(_a)) os << *it++;
  while (it != end(_a))
    os << "\n "[is_fundamental<typename T::value_type>::value] << *it++;
  return os;
}
template <class T1, class T2>
inline ostream& operator<<(ostream& os, pair<T1, T2> const& _a) {
  return os << _a.first << " " << _a.second;
}
template <class T1, class T2>
inline istream& operator>>(istream& is, pair<T1, T2>& _a) {
  return is >> _a.first >> _a.second;
}
template <class T>
inline istream& operator>>(istream& is, vector<T>& _a) {
  for (int i = 0; i < (int)(_a.size()); ++i) is >> _a[i];
  return is;
}
template <class T>
inline void print(T&& _a) {
  cout << " " << _a;
}
template <class T>
inline void priws(T&& _a) {
  cout << _a;
}
template <class T>
inline void read(T& _a) {
  cin >> _a;
}
