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 power(long long int a, long long int b) {
  long long int res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long int modInverse(long long int a) { return power(a, 1000000007 - 2); }
const int N = 500023;
bool vis[N];
vector<int> adj[N];
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int my_a = 0;
  int mx_len = INT_MIN;
  for (long long int q = 0; q < 26; q++) {
    my_a++;
    for (long long int i = 0; i < n; i++) {
      int cnt1 = 0, cnt2 = 0;
      int mx = INT_MIN;
      for (long long int j = 0; j < i; j++) {
        if (a[j] == my_a) {
          cnt1++;
        }
      }
      int j = n - 1;
      while (cnt2 < cnt1 && j >= i) {
        if (a[j] == my_a) {
          cnt2++;
        }
        j--;
      }
      if (cnt1 == cnt2) {
        int freq[30];
        for (long long int k = 0; k < 30; k++) freq[k] = 0;
        for (int k = i; k < j + 1; k++) {
          freq[a[k]]++;
        }
        for (long long int k = 0; k < 30; k++) {
          mx = max(freq[k], mx);
        }
      }
      mx_len = max(mx + cnt1 + cnt2, mx_len);
    }
  }
  cout << mx_len << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  cin >> T;
  int t = 0;
  while (t++ < T) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}
