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 long long N = 2e5 + 7;
const long long MOD = 1e9 + 7;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long> v(n);
    map<long long, vector<long long> > m;
    long long mx = 0, id;
    for (long long i = 0; i < n; i++) {
      cin >> v[i];
      m[v[i]].push_back(i);
      mx = max(mx, (long long)m[v[i]].size());
    }
    long long res = mx;
    for (auto it : m) {
      long long sz = it.second.size(), ans = 0;
      long long cnt = 0, cnt2 = sz - 1, cntf = 0;
      long long i = it.second[cnt], j = it.second[cnt2];
      while (cnt < cnt2) {
        ans = 0;
        mx = 0;
        cntf += 2;
        map<long long, long long> temp;
        for (long long k = i + 1; k < j; k++) {
          if (v[k] == it.first) continue;
          temp[v[k]]++;
          mx = max(mx, temp[v[k]]);
        }
        ans = max(ans, mx) + cntf;
        cnt++, cnt2--;
        i = it.second[cnt], j = it.second[cnt2];
        res = max(res, ans);
      }
    }
    cout << res << '\n';
  }
  return 0;
}
