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;
long long int b[27][2009];
long long int wow(long long int e, long long int l, long long int r) {
  if (r < l) return 0;
  return b[e][r] - b[e][l - 1];
}
int main() {
  long long int t;
  cin >> t;
  for (long long int i = 0; i <= 26; i++) b[i][0] = 0;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n + 1];
    for (long long int i = 1; i <= n; i++) {
      cin >> a[i];
      for (long long int j = 1; j <= 26; j++) {
        b[j][i] = 0;
        if (j == a[i]) b[j][i] = 1;
        b[j][i] += b[j][i - 1];
      }
    }
    long long int ans = 0;
    for (long long int i = 1; i <= n; i++) {
      for (long long int j = i; j <= n; j++) {
        long long int m1 = 0, m2 = 0;
        for (long long int k = 1; k <= 26; k++) {
          m1 = max(m1, wow(k, i, j));
          m2 = max(m2, min(wow(k, 1, i - 1), wow(k, j + 1, n)));
        }
        ans = max(ans, m2 + m1 + m2);
      }
    }
    cout << ans << endl;
  }
}
