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];
    int b[n];
    int i, j;
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    int left[n][27];
    int right[n][27];
    memset(left, 0, sizeof(left));
    memset(right, 0, sizeof(right));
    left[0][a[0]] = 1;
    for (i = 1; i < n; i++) {
      for (j = 1; j <= 26; j++) {
        if (j == a[i]) {
          left[i][j] = left[i - 1][j] + 1;
        } else {
          left[i][j] = left[i - 1][j];
        }
      }
    }
    right[n - 1][a[n - 1]] = 1;
    for (i = n - 2; i >= 0; i--) {
      for (j = 1; j <= 26; j++) {
        if (j == a[i]) {
          right[i][j] = right[i + 1][j] + 1;
        } else {
          right[i][j] = right[i + 1][j];
        }
      }
    }
    int ans = 1;
    for (i = 0; i < n; i++) {
      for (j = n - 1; j >= 0; j--) {
        if (i >= j) {
          continue;
        }
        for (int k = 1; k <= 26; k++) {
          if (left[i][k] == right[j][k]) {
            for (int l = 1; l <= 26; l++) {
              ans = max(ans, 2 * left[i][k] + left[j - 1][l] - left[i][l]);
            }
          }
        }
      }
    }
    cout << ans << endl;
  }
}
