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 int N = 200000 + 200;
int cnt[N];
int a[N];
int n;
int calc(int x) {
  for (int i = 1; i <= 200; i++) cnt[i] = 0;
  for (int i = 1; i <= n; i++) cnt[a[i]]++;
  priority_queue<pair<int, int> > que;
  int ans = 0;
  for (int i = 1; i <= 200; i++) {
    que.push({cnt[i], i});
    ans = max(ans, cnt[i]);
  }
  int L = 1, R = n;
  int t = 0;
  while (L < R) {
    while (L <= n && a[L] != x) {
      cnt[a[L]]--;
      L++;
    }
    while (R > 0 && a[R] != x) {
      cnt[a[R]]--;
      R--;
    }
    if (L >= R) break;
    cnt[x] -= 2;
    L++, R--;
    t += 2;
    while (cnt[que.top().second] != que.top().first) {
      int y = que.top().second;
      que.pop();
      que.push({cnt[y], y});
    }
    ans = max(ans, t + que.top().first);
  }
  return ans;
}
void solve() {
  int ans = 0;
  for (int i = 1; i <= 200; i++) {
    ans = max(ans, calc(i));
  }
  printf("%d\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    solve();
  }
  return 0;
}
