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;
using ll = long long;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
void ez() {
  int n;
  cin >> n;
  vector<int> a(n);
  int cnt[30][n + 1];
  for (int i = 0; i < (30); i++)
    for (int j = 0; j < (n + 1); j++) cnt[i][j] = 0;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    for (int j = 0; j < (30); j++) {
      cnt[j][i + 1] = cnt[j][i];
    }
    cnt[a[i]][i + 1]++;
  }
  int ans = 0;
  for (int i = 0; i < (n + 1); i++) {
    for (int j = i; j <= n; j++) {
      int x = 0, y = 0;
      for (int k = 0; k < (30); k++) {
        chmax(x, min(cnt[k][i], cnt[k][n] - cnt[k][j]));
        chmax(y, cnt[k][j] - cnt[k][i]);
      }
      chmax(ans, x * 2 + y);
    }
  }
  cout << ans << endl;
}
void hd() {}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int q;
  cin >> q;
  while (q--) ez();
  return 0;
}
