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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[n + 1][200];
    memset(a, 0, sizeof(a));
    vector<int> v[200];
    for (long long i = 1; i < n + 1; i++) {
      int x;
      cin >> x;
      x--;
      a[i][x]++;
      v[x].push_back(i);
    }
    for (long long i = 0; i < 200; i++) {
      for (long long j = 1; j < n + 1; j++) a[j][i] += a[j - 1][i];
    }
    int ans = -1;
    for (long long i = 0; i < 200; i++) {
      ans = max(ans, (int)v[i].size());
      for (long long j = 0; j < v[i].size() / 2; j++) {
        int l = v[i][j] + 1, r = v[i][v[i].size() - j - 1] - 1;
        int val = -1;
        for (long long k = 0; k < 200; k++) {
          val = max(val, a[r][k] - a[l - 1][k]);
        }
        ans = max((long long)ans, val + 2 * (j + 1));
      }
    }
    cout << ans << "\n";
  }
}
