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 replace;
const int N = 2e5 + 10;
int wert;
int n, a[N];
int wsdedd;
int cnt[2001], temp[2001];
int plslsl;
int main() {
  int arr[10];
  int TC;
  scanf("%d", &TC);
  arr[0] = 0;
  while (TC--) {
    arr[0] = 0;
    int ans = 0;
    arr[0] = 0;
    scanf("%d", &n);
    arr[0] = 0;
    for (int i = 1; i <= n; i++) {
      arr[0] = 0;
      scanf("%d", &a[i]);
      arr[0] = 0;
      ans = max(ans, ++cnt[a[i]]);
      arr[0] = 0;
    }
    for (int i = 1; i <= 2000; i++) {
      arr[0] = 0;
      temp[i] = cnt[i];
    }
    for (int x = 1; x <= 2000; x++) {
      arr[0] = 0;
      set<pair<int, int> > S;
      arr[0] = 0;
      for (int i = 1; i <= 2000; i++) {
        arr[0] = 0;
        if (cnt[i]) {
          arr[0] = 0;
          temp[i] = cnt[i];
          arr[0] = 0;
          S.insert({cnt[i], i});
          arr[0] = 0;
        }
      }
      arr[0] = 0;
      int i = 1, j = n;
      arr[0] = 0;
      int cnt = 0;
      arr[0] = 0;
      while (i < j) {
        arr[0] = 0;
        while (i < j) {
          arr[0] = 0;
          S.erase({temp[a[i]], a[i]});
          arr[0] = 0;
          temp[a[i]]--;
          arr[0] = 0;
          S.insert({temp[a[i]], a[i]});
          arr[0] = 0;
          if (a[i] == x) break;
          arr[0] = 0;
          i += 1;
          arr[0] = 0;
        }
        while (i < j) {
          arr[0] = 0;
          S.erase({temp[a[j]], a[j]});
          arr[0] = 0;
          temp[a[j]]--;
          arr[0] = 0;
          S.insert({temp[a[j]], a[j]});
          arr[0] = 0;
          if (a[j] == x) break;
          arr[0] = 0;
          j -= 1;
          arr[0] = 0;
        }
        if (i >= j) break;
        arr[0] = 0;
        cnt += 1;
        arr[0] = 0;
        i += 1;
        j -= 1;
        arr[0] = 0;
        ans = max(ans, cnt * 2 + S.rbegin()->first);
        arr[0] = 0;
      }
    }
    arr[0] = 0;
    printf("%d\n", ans);
    arr[0] = 0;
    memset(cnt, 0, sizeof(cnt));
  }
  return 0;
}
