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 int Mod = (long long int)1e9 + 7;
long long int CaseN = 1;
const long double PI = 22 / 7.0;
long long int tst() {
  long long int ans = 0;
  return ans;
}
void solve() {
  long long int n;
  cin >> n;
  long long int p[202][n + 1];
  for (long long int i = 1; i <= 200; i++) {
    memset(p[i], 0, sizeof(p[i]));
  }
  vector<long long int> occ(201, 0);
  for (long long int i = 1; i <= n; i++) {
    long long int a;
    cin >> a;
    occ[a]++;
    p[a][occ[a]] = i;
  }
  long long int ans = 0;
  for (long long int i = 1; i <= 200; i++) {
    long long int k = occ[i];
    for (long long int j = 1; j <= k / 2; j++) {
      long long int l = p[i][j] + 1;
      long long int r = p[i][k - j + 1] - 1;
      long long int cnt1 = j * 2;
      long long int cnt2 = 0;
      if (l <= r) {
        for (long long int o = 1; o <= 200; o++) {
          long long int in1 =
              upper_bound(p[o] + 1, p[o] + 1 + occ[o], r) - 1 - p[o];
          long long int in2 =
              lower_bound(p[o] + 1, p[o] + 1 + occ[o], l) - p[o];
          cnt2 = max(cnt2, in1 - in2 + 1);
        }
      }
      ans = max(ans, (cnt1 + cnt2));
    }
  }
  if (ans == 0) {
    ans = 1;
  }
  cout << ans << endl;
  CaseN++;
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
