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;
const long long N = 2e5 + 5;
const long long types = 30;
long long mas[N];
long long pref[N][types];
long long suf[N][types];
long long sum(long long l, long long r, long long type) {
  return pref[r][type] - (l ? pref[l - 1][type] : 0);
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> mas[i];
  }
  long long ans = 1;
  for (long long i = 0; i < n; i++) {
    for (long long type = 0; type < types; type++) {
      pref[i][type] = (mas[i] == type);
      if (i) {
        pref[i][type] += pref[i - 1][type];
      }
    }
  }
  for (long long i = (long long)n - 1; i >= 0; --i) {
    for (long long type = 0; type < types; type++) {
      suf[i][type] = (mas[i] == type);
      if (i != n - 1) {
        suf[i][type] += suf[i + 1][type];
      }
    }
  }
  for (long long l = 0; l < n; l++) {
    for (long long r = l + 1; r < n; r++) {
      long long mxX = 0;
      for (long long type = 0; type < types; type++) {
        long long x = min(pref[l][type], suf[r][type]);
        mxX = max(x, mxX);
      }
      long long mxY = 0;
      for (long long type = 0; type < types; type++) {
        long long y = sum(l + 1, r - 1, type);
        mxY = max(mxY, y);
      }
      ans = max(ans, mxX * 2 + mxY);
    }
  }
  cout << ans;
}
int main() {
  long long tests = 1;
  cin >> tests;
  while (tests--) {
    solve();
    cout << "\n";
  }
  return 0;
}
