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>
#pragma GCC optimize "03"
using namespace std;
const long long int inf = 1e15;
const long long int MOD = 1e9 + 7;
const long long int N = 2005;
long long int a[N];
long long int f[N][30], b[N][30];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    for (long long int i = 1; i <= n; i++) {
      cin >> a[i];
      for (long long int j = 1; j <= 26; j++) {
        f[i][j] = f[i - 1][j];
      }
      f[i][a[i]]++;
    }
    for (long long int i = n; i >= 1; i--) {
      for (long long int j = 1; j <= 26; j++) {
        b[i][j] = b[i + 1][j];
      }
      b[i][a[i]]++;
    }
    long long int ans = 0;
    for (long long int i = 1; i <= n; i++) {
      for (long long int j = i + 1; j <= n; j++) {
        long long int dis = 0;
        for (long long int k = 1; k <= 26; k++) {
          dis = max(dis, f[j - 1][k] - f[i][k]);
        }
        long long int cnt = 0;
        for (long long int k = 1; k <= 26; k++) {
          cnt = max(cnt, min(b[j][k], f[i][k]));
        }
        ans = max(ans, cnt * 2 + dis);
      }
    }
    if (ans == 0)
      cout << 1 << '\n';
    else
      cout << ans << '\n';
    for (long long int i = 1; i <= n; i++) {
      for (long long int j = 1; j <= 26; j++) {
        b[i][j] = f[i][j] = 0;
      }
    }
  }
  return 0;
}
