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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t-- > 0) {
    long long n;
    cin >> n;
    long long a[n];
    long long i, j, k;
    for (i = 0; i < n; i++) cin >> a[i];
    unordered_map<long long, long long> m, m2;
    long long ans = 0;
    long long dp[n][27];
    vector<long long> v[27];
    memset(dp, 0ll, sizeof(dp));
    for (i = 0; i < n; i++) {
      v[a[i]].push_back(i + 1);
      if (i == 0) {
        dp[i][a[i]]++;
      } else {
        for (j = 1; j <= 26; j++) {
          dp[i][j] += dp[i - 1][j];
        }
        dp[i][a[i]]++;
      }
    }
    for (i = 1; i < 27; i++) {
      if (v[i].size() == 0) continue;
      long long q = v[i].size();
      for (j = 0; j < q / 2; j++) {
        long long x = v[i][j];
        long long y = v[i][q - j - 1];
        x--;
        y--;
        for (k = 1; k < 27; k++) {
          if (k == i) continue;
          long long l = dp[y][k] - dp[x][k];
          ans = max(ans, l + 2 * (j + 1));
        }
        ans = max(ans, (long long)v[i].size());
      }
    }
    if (ans == 0) ans = 1;
    cout << ans << endl;
  }
}
