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 MAXN = 2010;
const int MAX_ALPHA = 26;
int n;
int a[MAXN];
int prefix[MAX_ALPHA][MAXN];
void test_case() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= MAX_ALPHA; i++) {
    prefix[i][0] = 0;
    for (int j = 1; j <= n; j++) {
      prefix[i][j] = (a[j] == i) + prefix[i][j - 1];
    }
    prefix[i][n + 1] = prefix[i][n];
  }
  int solution = -1;
  for (int l = 1; l <= n; l++) {
    for (int r = l; r <= n; r++) {
      int inner_c = 0;
      for (int i = 1; i <= MAX_ALPHA; i++) {
        int count = prefix[i][r] - prefix[i][l - 1];
        inner_c = max(inner_c, count);
      }
      int outer_c = 0;
      for (int i = 1; i <= MAX_ALPHA; i++) {
        int count = min(prefix[i][l - 1], prefix[i][n + 1] - prefix[i][r]);
        outer_c = max(outer_c, count);
      }
      solution = max(solution, outer_c + inner_c + outer_c);
    }
  }
  printf("%d\n", solution);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    test_case();
  }
}
