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 = 2 * 1e3 + 3;
int n;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  for (int qq = 1; qq <= tt; ++qq) {
    cin >> n;
    vector<int> a(n + 1, 0);
    vector<vector<int>> mx(n + 1, vector<int>(n + 1, 0));
    vector<vector<int>> L(n + 2, vector<int>(27, 0));
    vector<vector<int>> R(n + 2, vector<int>(27, 0));
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int len = 1; len <= n; ++len) {
      vector<int> cnt(27, 0);
      for (int i = 1; i <= len; ++i) {
        cnt[a[i]]++;
      }
      int cur = 0;
      for (int j = 1; j <= 26; ++j) cur = max(cur, cnt[j]);
      mx[1][len] = cur;
      for (int i = len + 1; i <= n; ++i) {
        cnt[a[i - len]]--;
        cnt[a[i]]++;
        cur = 0;
        for (int j = 1; j <= 26; ++j) cur = max(cur, cnt[j]);
        mx[i - len + 1][i] = cur;
      }
    }
    for (int d = 1; d <= 26; ++d) {
      for (int i = 1; i <= n; ++i) {
        if (a[i] == d) L[i][d]++;
        L[i][d] += L[i - 1][d];
      }
      for (int i = n; i >= 1; i--) {
        if (a[i] == d) R[i][d]++;
        R[i][d] += R[i + 1][d];
      }
    }
    long long res = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = i + 1; j <= n; ++j) {
        for (int d = 1; d <= 26; ++d) {
          int tl = L[i][d], tr = R[j][d];
          if (tl != tr) continue;
          int bw = mx[i + 1][j - 1];
          res = max(res, (long long)(tl + tr + bw));
        }
      }
    }
    if (res == 0) res = 1;
    cout << res << '\n';
  }
  return 0;
}
