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;
const long long N = 2005;
long long a[N], l[N][30], r[N][30];
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i <= n + 1; i++)
    for (long long j = 1; j <= 26; j++) {
      l[i][j] = 0;
      r[i][j] = 0;
    }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= 26; j++) {
      l[i][j] = l[i - 1][j];
    }
    l[i][a[i]]++;
  }
  for (long long i = n; i >= 1; i--) {
    for (long long j = 1; j <= 26; j++) {
      r[i][j] = r[i + 1][j];
    }
    r[i][a[i]]++;
  }
  long long ans = 0;
  for (long long i = 0; i <= n; i++)
    for (long long j = i + 1; j <= n + 1; j++)
      for (long long k = 1; k <= 26; k++)
        for (long long m = 1; m <= 26; m++) {
          ans = max(ans, min(l[i][k], r[j][k]) * 2 + l[j - 1][m] - l[i][m]);
        }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) solve();
}
