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;
bool check(int n, int pos) { return n & (1LL << pos); }
int Set(int n, int pos) { return n = n | (1LL << pos); }
int arr[200005];
int freq[203][200005];
int main() {
  int i, j, k, l, m, n, o, r, q;
  int testcase;
  int input, flag, tag, ans;
  scanf("%d", &testcase);
  for (int cs = 1; cs <= testcase; cs++) {
    scanf("%d", &n);
    for (i = 1; i <= 200; i++) {
      for (j = 1; j <= n; j++) {
        freq[i][j] = 0;
      }
    }
    ans = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &input);
      arr[i] = input;
      freq[input][i]++;
    }
    for (i = 1; i <= 200; i++) {
      for (j = 1; j <= n; j++) {
        freq[i][j] += freq[i][j - 1];
      }
      ans = max(ans, freq[i][n]);
    }
    for (i = 1; i <= 200; i++) {
      int p1, p2;
      p1 = 1;
      p2 = n;
      int cnt = 0;
      while (p1 < p2) {
        while (arr[p1] != i && p1 <= p2) p1++;
        while (arr[p2] != i && p1 <= p2) p2--;
        cnt += 2;
        if (p1 >= p2) break;
        for (j = 1; j <= 200; j++) {
          if (i == j) continue;
          int cnt2 = 0;
          cnt2 = freq[j][p2 - 1] - freq[j][p1];
          ans = max(ans, cnt2 + cnt);
        }
        p1++;
        p2--;
      }
    }
    printf("%lld\n", ans);
  }
}
