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;
int ans = 0;
vector<int> v;
void mxpolindrome() {
  int l = 0, r = v.size() - 1;
  ;
  int mid = 0;
  for (auto i : v) mid += i;
  int per = 0;
  while (l <= r) {
    per++;
    bool l0 = 0, r0 = 0;
    while (!l0) {
      if (!v[l])
        l0 = 1;
      else
        mid--;
      l++;
    }
    while (!r0) {
      if (!v[r])
        r0 = 1;
      else
        mid--;
      r--;
    }
    if (!r0 || !l0 || r < l) break;
    ans = max(ans, 2 * per + mid);
  }
}
void solve() {
  int n;
  cin >> n;
  ans = 0;
  map<int, vector<int>> ind;
  for (int i = (int)(0); i < (int)(n); i++) {
    int a;
    cin >> a;
    ind[a].push_back(i);
  }
  for (int i = (int)(1); i < (int)(prev(ind.end())->first + 1); i++) {
    if (ind[i].size() == 0) continue;
    for (int j = (int)(1); j < (int)(prev(ind.end())->first + 1); j++) {
      if (j == i || ind[j].size() == 0) {
        ans = max(ans, (int)ind[i].size());
        continue;
      }
      for (int p = 0, l = 0; p < ind[i].size() || l < ind[j].size();) {
        if (l == ind[j].size()) {
          v.push_back(0), p++;
          continue;
        }
        if (p == ind[i].size()) {
          v.push_back(1), l++;
          continue;
        }
        if (ind[i][p] < ind[j][l])
          v.push_back(0), p++;
        else
          v.push_back(1), l++;
      }
      mxpolindrome();
      v.clear();
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tt;
  cin >> tt;
  while (tt--) solve();
}
