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 MAX = 2e3 + 4;
int n, cnt[27], ans;
vector<int> a, pf[27], sf[27];
void init() {
  memset(cnt, 0, sizeof(cnt));
  ans = 0;
  a.clear();
  a.resize(n + 1);
  for (int i = 0; i <= 26; i++) {
    pf[i].clear();
    sf[i].clear();
    pf[i].resize(n + 2);
    sf[i].resize(n + 2);
  }
}
void solve() {
  cin >> n;
  init();
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
    ans = max(ans, cnt[a[i]]);
    for (int k = 1; k <= 26; k++) pf[k][i] = pf[k][i - 1] + (k == a[i]);
  }
  for (int i = n; i >= 1; i--)
    for (int k = 1; k <= 26; k++) sf[k][i] = sf[k][i + 1] + (k == a[i]);
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      int t1 = 0, t2 = 0;
      for (int k = 1; k <= 26; k++) {
        t1 = max(t1, pf[k][j] - pf[k][i - 1]);
      }
      for (int k = 1; k <= 26; k++) {
        t2 = max(t2, min(pf[k][i - 1], sf[k][j + 1]));
      }
      ans = max(ans, t1 + 2 * t2);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  int tt;
  cin >> tt;
  while (tt--) solve();
}
