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;
long long int recpow(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return recpow((x * x) % 1000000007, n / 2);
  else
    return (x * recpow((x * x) % 1000000007, (n - 1) / 2)) % 1000000007;
}
bool sortbysec(
    const pair<pair<long long int, long long int>, long long int> &a,
    const pair<pair<long long int, long long int>, long long int> &b) {
  if (a.second == b.second)
    return (a.first.second < b.first.second);
  else
    return (a.second > b.second);
}
bool comp(pair<long long int, long long int> a,
          pair<long long int, long long int> b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first < b.first;
}
string rev(string s) {
  string t;
  for (long long int i = s.length() - 1; i >= 0; i--) {
    t.push_back(s[i]);
  }
  return t;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    long long int f[n][27], g[n][27];
    for (long long int i = 0; i < n; i++)
      for (long long int j = 0; j < 27; j++) f[i][j] = 0, g[i][j] = 0;
    cin >> a[0];
    f[0][a[0]]++;
    for (long long int i = 1; i < n; i++) {
      cin >> a[i];
      for (long long int j = 0; j < 27; j++) {
        f[i][j] = f[i - 1][j];
      }
      f[i][a[i]]++;
    }
    g[n - 1][a[n - 1]]++;
    for (long long int i = n - 2; i >= 0; i--) {
      for (long long int j = 0; j < 27; j++) {
        g[i][j] = g[i + 1][j];
      }
      g[i][a[i]]++;
    }
    long long int ans = 0;
    for (long long int i = 0; i < n; i++) {
      for (long long int j = n - 1; j >= i; j--) {
        if (f[i][a[i]] == g[j][a[i]]) {
          if (i < j) {
            for (long long int k = 0; k < 27; k++) {
              if (k != a[i])
                ans = max(ans, f[i][a[i]] + g[j][a[i]] + f[j][k] - f[i][k]);
              else
                ans = max(ans, f[i][a[i]] + g[j][a[i]] + f[j][k] - f[i][k] - 1);
            }
          } else {
            ans = max(ans, f[i][a[i]] + g[j][a[i]] - 1);
          }
          break;
        }
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
