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 a[2010];
int cnt[30];
int pref[2010][30];
int n;
int calc(int prvi, int drugi, int p1, int p2) {
  if (pref[p1 - 1][prvi] - pref[0][prvi] != pref[n][prvi] - pref[p2][prvi])
    return false;
  return pref[p1 - 1][prvi] - pref[0][prvi] + pref[n][prvi] - pref[p2][prvi] +
         pref[p2][drugi] - pref[p1 - 1][drugi];
}
int izracunaj(int p1, int p2) {
  int rez = 0;
  for (int i = 1; i <= 26; i++) {
    for (int j = 1; j <= 26; j++) {
      if (i == j) continue;
      bool ok = false;
      if (rez < 26) ok = true;
      rez = max(rez, calc(i, j, p1, p2));
    }
  }
  return rez;
}
void resi() {
  memset(cnt, 0, sizeof(cnt));
  memset(pref, 0, sizeof(pref));
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
    for (int j = 1; j <= 26; j++) pref[i][j] = pref[i - 1][j];
    pref[i][a[i]]++;
  }
  int rez = 0;
  for (int i = 1; i <= 26; i++) rez = max(rez, cnt[i]);
  for (int p1 = 2; p1 < n; p1++) {
    for (int p2 = p1; p2 < n; p2++) {
      rez = max(rez, izracunaj(p1, p2));
    }
  }
  cout << rez << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) resi();
  return 0;
}
