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>
int pref[26][2005];
int sum(int c, int l, int r) {
  if (l > r) return 0;
  if ((l < 0) || (r < 0))
    return 0;
  else
    return (pref[c][r + 1] - pref[c][l]);
}
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    int i, j, k, n;
    cin >> n;
    for (i = 0; i < 26; i++) {
      for (j = 0; j < 2005; j++) {
        pref[i][j] = 0;
      }
    }
    int a[n];
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (i = 0; i < 26; i++) {
      for (j = 1; j <= n; j++) {
        if (a[j - 1] == i + 1)
          pref[i][j] = pref[i][j - 1] + 1;
        else
          pref[i][j] = pref[i][j - 1];
      }
    }
    int ans = pref[0][n];
    for (i = 1; i < 26; i++) {
      ans = max(ans, pref[i][n]);
    }
    for (i = 0; i < n; i++) {
      for (j = i; j < n; j++) {
        int c1 = 0, c2 = 0;
        for (k = 0; k < 26; k++) {
          c1 = max(c1, sum(k, i, j));
        }
        for (k = 0; k < 26; k++) {
          c2 = max(c2, min(sum(k, 0, i - 1), sum(k, j + 1, n - 1)));
        }
        ans = max(ans, c1 + 2 * c2);
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
