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 a[200005], n;
int ma[201];
int Cnt[201][200005];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    int res = 0;
    memset(ma, 0, sizeof ma);
    set<int> st;
    vector<int> Pos[201];
    for (int j = 1; j <= n; j++) {
      ma[a[j]]++;
      st.insert(a[j]);
      Pos[a[j]].push_back(j);
      Cnt[a[j]][j]++;
    }
    for (auto x : st) {
      for (int i = 1; i <= n; i++) {
        Cnt[x][i] += Cnt[x][i - 1];
      }
      res = max(Cnt[x][n], res);
    }
    for (auto i : st) {
      int lastL = 1, lastR = n;
      int cntL = 0, cntR = 0;
      for (int j = 0; j < Pos[i].size() / 2; j++) {
        int lpos = Pos[i][j];
        int rpos = Pos[i][Pos[i].size() - 1 - j] - 1;
        for (auto k : st) {
          res = max(res, 2 * (j + 1) + Cnt[k][rpos] - Cnt[k][lpos]);
        }
      }
    }
    for (auto x : st) {
      for (int i = 1; i <= n; i++) {
        Cnt[x][i] = 0;
      }
    }
    printf("%d\n", res);
  }
  return 0;
}
