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 N = (int)2e3 + 10;
const int mod = 1e9 + 7;
long long n, ans, cnt;
long long dp[N][27];
int b[N];
vector<int> a[27];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    int g = 1;
    ans = 0;
    for (int i = 1; i <= n; i++) memset(dp[i], 0, sizeof(dp[i]));
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      for (int j = 1; j <= 26; j++) dp[i][j] = dp[i - 1][j];
      dp[i][x]++;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= 26; j++) {
        if (dp[i][j] != 0) {
          int ind = 0;
          for (int k = n; k > i; k--)
            if (dp[i][j] == (dp[n][j] - dp[k - 1][j])) {
              ind = k;
              break;
            }
          if (ind != 0) {
            int ma = 0;
            for (int u = 1; u <= 26; u++)
              ma = ma > (dp[ind - 1][u] - dp[i][u])
                       ? ma
                       : (dp[ind - 1][u] - dp[i][u]);
            ans = (ans > dp[i][j] * 2 + ma) ? ans : dp[i][j] * 2 + ma;
          }
        }
      }
    }
    for (int i = 1; i <= 26; i++) ans = ans > dp[n][i] ? ans : dp[n][i];
    cout << ans << endl;
  }
  return 0;
}
