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 modex(long long a, long long b) {
  long long ans = 1;
  a %= 1000000007;
  ;
  if (a == 0) return 0;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % 1000000007;
    ;
    b = b >> 1;
    a = (a * a) % 1000000007;
    ;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  int n;
  cin >> n;
  int a[n];
  int count[n][27];
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < 27; ++j) count[i][j] = 0;
  }
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    if (i) {
      for (long long j = 0; j < 27; ++j) {
        count[i][j] = count[i - 1][j];
      }
    }
    count[i][a[i]]++;
  }
  int ans = 0;
  for (long long i = 0; i < n; ++i) {
    for (long long j = n - 1; j >= i + 1; --j) {
      if (a[i] != a[j]) continue;
      for (long long k = 1; k < 27; ++k) {
        if (k == a[i]) continue;
        int t = count[j][k] - count[i][k];
        int first = count[i][a[i]], second = 0;
        if (j == n - 1)
          second = 1;
        else
          second = (count[n - 1][a[i]] - count[j][a[i]] + 1);
        ans = max(ans, t + 2 * min(first, second));
      }
    }
  }
  for (long long i = 0; i < 27; ++i) {
    ans = max(ans, count[n - 1][i]);
  }
  if (!ans) ans = 1;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  cin >> q;
  while (q--) solve();
}
