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;
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
const int nax = 2123;
int N;
int f[nax][26];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  for (int qq = 1; qq <= tt; ++qq) {
    memset(f, 0, sizeof(f));
    cin >> N;
    for (int i = 0; i < N; ++i) {
      int a;
      cin >> a;
      ++f[i][a - 1];
    }
    for (int i = 1; i < N; ++i) {
      for (int j = 0; j < 26; ++j) {
        f[i][j] += f[i - 1][j];
      }
    }
    int max_len = 0;
    for (int i = 0; i < 26; ++i) uax(max_len, f[N - 1][i]);
    for (int l = 0; l < N - 2; ++l) {
      for (int r = l + 1; r < N - 1; ++r) {
        int lr = 0, mid = 0;
        for (int a = 0; a < 26; ++a) {
          uax(lr, min(f[l][a], f[N - 1][a] - f[r][a]) * 2);
          uax(mid, f[r][a] - f[l][a]);
        }
        uax(max_len, lr + mid);
      }
    }
    cout << max_len << '\n';
  }
}
