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 main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[n + 5];
    for (int i = 0; i < n; i++) cin >> a[i];
    int ans = 1;
    for (int i = 1; i <= 26; i++) {
      int x = 1;
      while (true) {
        int count = 0;
        int j, k;
        for (j = 0; j < n; j++) {
          if (a[j] == i) count++;
          if (count == x) break;
        }
        int count1 = 0;
        for (k = n - 1; k >= 0; k--) {
          if (a[k] == i) count1++;
          if (count1 == x) break;
        }
        if (j == n || k == -1) break;
        if (j >= k) break;
        vector<int> freq;
        for (int l = 0; l <= 27; l++) freq.push_back(0);
        for (int l = j + 1; l < k; l++) freq[a[l]]++;
        int y = 0;
        for (int l = 0; l <= 26; l++) y = max(y, freq[l]);
        ans = max(ans, 2 * x + y);
        x++;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
