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 = 2000;
int v[maxn], pos[27][maxn], cnt[27];
void init(int);
int main(void) {
  int t;
  cin >> t;
  while (t--) {
    int len;
    cin >> len;
    for (int i = 0; i < len; i++) cin >> v[i];
    init(len);
    int res = 0;
    for (int i = 1; i <= 26; i++)
      for (int posl = 0, posr = cnt[i] - 1; posl <= posr; posl++, posr--)
        for (int j = 1; j <= 26; j++) {
          int left =
              lower_bound(pos[j], pos[j] + cnt[j], pos[i][posl]) - pos[j];
          int right =
              lower_bound(pos[j], pos[j] + cnt[j], pos[i][posr]) - pos[j];
          int val = (posl + 1) * 2 + right - left;
          if (posl == posr) val--;
          if (i == j) val = cnt[i];
          res = max(res, val);
        }
    cout << res << endl;
  }
  return 0;
}
void init(int len) {
  memset(cnt, 0, sizeof(cnt));
  memset(pos, 0, sizeof(pos));
  for (int i = 0; i < len; i++) pos[v[i]][cnt[v[i]]++] = i;
}
