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;
int maxi;
int n, arr[2005];
void func(int a, int b) {
  int i, j;
  int pre[2005];
  memset(pre, 0, sizeof(pre));
  pre[0] = 0;
  for (i = 1; i <= n; i++) {
    if (arr[i] == b)
      pre[i] = pre[i - 1] + 1;
    else
      pre[i] = pre[i - 1];
  }
  int s = 0;
  int e = n + 1;
  int coun = 0;
  while (s < e) {
    for (i = s + 1; i <= e; i++)
      if (arr[i] == a) break;
    for (j = e - 1; j >= s; j--)
      if (arr[j] == a) break;
    if (i >= j)
      break;
    else {
      coun = coun + 2;
      maxi = max(maxi, coun + pre[j] - pre[i - 1]);
      s = i;
      e = j;
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int i, j;
    cin >> n;
    for (i = 1; i <= n; i++) cin >> arr[i];
    int freq[27] = {};
    for (i = 1; i <= n; i++) freq[arr[i]]++;
    maxi = 0;
    for (i = 1; i <= 26; i++)
      for (i = 1; i <= 26; i++) maxi = max(maxi, freq[i]);
    for (i = 1; i <= 25; i++)
      for (j = i + 1; j <= 26; j++) {
        if (freq[i] != 0 && freq[j] != 0) {
          func(i, j);
          func(j, i);
        }
      }
    cout << maxi << endl;
  }
  return 0;
}
