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 = 3e5 + 500;
const long long mod = 1e9 + 7;
const long long cmod = 998244353;
const long long inf = 1LL << 61;
const int M = 1e6 + 500;
const long long int ths = 1LL << 40;
const int NN = 5e3 + 6;
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int pre[n][207], suf[207][n + 10];
  memset(pre, 0, sizeof pre);
  memset(suf, 0, sizeof suf);
  for (int i = 0; i < n; i++) {
    if (i != 0) {
      for (int j = 1; j <= 26; j++) {
        pre[i][j] = pre[i - 1][j];
      }
    }
    pre[i][a[i]]++;
  }
  long long int f[207] = {};
  for (int i = n - 1; i >= 0; i--) {
    f[a[i]]++;
    suf[a[i]][f[a[i]]] = i;
  }
  long long int ans = 0;
  for (int i = 1; i <= 200; i++) {
    ans = max(ans, pre[n - 1][i]);
    for (int j = 0; j < n; j++) {
      long long int l = j;
      long long int cnt = pre[j][i];
      long long int r = suf[i][cnt];
      if (r <= l) continue;
      for (int k = 1; k <= 200; k++) {
        long long int x = pre[r - 1][k] - pre[l][k];
        long long int t = cnt * 2 + x;
        ans = max(ans, cnt * 2 + x);
      }
    }
  }
  cout << ans;
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
