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 n;
long long a[3000];
long long pre[30][3000], suf[30][3000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int i, t, j;
  cin >> t;
  while (t--) {
    cin >> n;
    for (i = 0; i <= n + 1; i++)
      for (j = 1; j <= 26; j++) pre[j][i] = 0, suf[j][i] = 0;
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= 26; j++) pre[j][i] = pre[j][i - 1] + (a[i] == j);
    }
    for (i = n; i > 0; i--) {
      for (j = 1; j <= 26; j++) suf[j][i] = suf[j][i + 1] + (a[i] == j);
    }
    for (i = 1; i <= 26; i++) {
      sort(suf[i] + 1, suf[i] + n + 1);
    }
    long long ans = 0;
    for (i = 1; i <= n; i++) {
      long long sum = 0;
      for (j = 1; j <= 26; j++) {
        sum = 0;
        long long r =
            lower_bound(suf[j] + 1, suf[j] + n + 1, pre[j][i]) - suf[j];
        long long u = n - r + 1;
        if (u > i) {
          long long mx2 = 0;
          sum = pre[j][i] * 2;
          for (long long k = 1; k <= 26; k++) {
            mx2 = max(mx2, pre[k][u - 1] - pre[k][i]);
          }
          sum += mx2;
        }
        ans = max(ans, sum);
      }
    }
    if (ans == 0) ans = 1;
    cout << ans << endl;
  }
  return 0;
}
