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 int N = 2e5 + 5;
int t, n;
int a[2005];
int cnt[30][2005];
int main() {
  cin >> t;
  while (t--) {
    scanf("%d", &n);
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      for (int j = 1; j <= 26; j++) {
        cnt[j][i] = cnt[j][i - 1];
      }
      cnt[a[i]][i]++;
    }
    int maxx = 1;
    for (int i = 1; i <= n; i++) {
      int maxx1 = 0, maxx2 = 0, maxx3 = 0;
      for (int z = 1; z <= 26; z++) {
        maxx1 = cnt[z][i];
        if (maxx1 == 0) continue;
        for (int j = i; j <= n; j++) {
          maxx2 = 0;
          for (int k = 1; k <= 26; k++) {
            maxx2 = cnt[k][j] - cnt[k][i];
            maxx3 = cnt[z][n] - cnt[z][j];
            if (maxx3 < maxx1) continue;
            maxx3 = maxx1;
            maxx = max(maxx, maxx1 + maxx2 + maxx3);
          }
        }
      }
    }
    printf("%d\n", maxx);
  }
}
