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;
long long t, n, m, r, l, h, i, j, ii, x[1000001], y[1000001], z, g, a, b, c, d,
    k, q;
vector<long long> v;
vector<pair<long long, long long> > vv;
queue<pair<long long, long long> > qq;
pair<long long, long long> p[1000001];
map<long long, long long> mp, mpp, mx, vis;
set<long long> se;
set<long long>::iterator it;
string s[1000001], sss;
char cc;
int main() {
  ios::sync_with_stdio(0);
  n = 9;
  cin >> t;
  while (t--) {
    cin >> n;
    z = 1;
    mp.clear();
    for (i = 0; i < n; i++) cin >> x[i];
    for (q = 0; q < n / 2; q++) {
      if (mp[x[q]] == 0) {
        m = x[q];
        j = n - q - 1;
        for (i = q; i < j; i++) {
          if (x[i] == m) {
            mp[x[i]]++;
            for (; j > i; j--) {
              if (x[j] == m) {
                mp[x[i]]++;
                a = 0;
                mpp.clear();
                for (k = i + 1; k < j; k++) {
                  mpp[x[k]]++;
                  a = max(a, mpp[x[k]]);
                }
                z = max(mp[m] + a, z);
                j--;
                break;
              }
            }
          }
        }
      }
      if (mp[x[n - q - 1]] == 0) {
        m = x[n - q - 1];
        j = q;
        for (i = n - q - 1; i > j; i--) {
          if (x[i] == m) {
            mp[x[i]]++;
            for (; j < i; j++) {
              if (x[j] == m) {
                mp[x[i]]++;
                a = 0;
                mpp.clear();
                for (k = j + 1; k < i; k++) {
                  mpp[x[k]]++;
                  a = max(a, mpp[x[k]]);
                }
                z = max(mp[m] + a, z);
                j++;
                break;
              }
            }
          }
        }
      }
    }
    cout << z;
    cout << endl;
  }
  return 0;
}
