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 long long oo = 1e8;
const double pi = 3.1415926535897;
const double EPS = (1e-7);
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
struct edge {
  int from, to, w;
  edge() {}
  edge(int x, int y, int we) : from(x), to(y), w(we) {}
  bool operator<(const edge &rhs) const { return w > rhs.w; }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int A[n + 5];
    int frq[n + 5][26 + 5];
    memset(frq, 0, sizeof(frq));
    for (int i = 1; i <= n; i++) {
      cin >> A[i];
      frq[i][A[i]]++;
      for (int j = 1; j <= 26; j++) {
        frq[i][j] += frq[i - 1][j];
      }
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      int x = frq[i][A[i]];
      for (int j = i + 1; j <= n; j++) {
        if ((frq[n][A[i]] - frq[j - 1][A[i]]) >= x) ans = max(ans, 2 * x);
        int y = frq[j][A[j]] - frq[i][A[j]];
        if (frq[n][A[i]] - frq[j][A[i]] >= x) ans = max(ans, 2 * x + y);
      }
    }
    for (int i = 1; i <= 26; i++) ans = max(ans, frq[n][i]);
    cout << ans << "\n";
  }
}
