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 double eps = 1e-9;
const long long inf = 1e8;
const int N = 2e3 + 5;
const int M = 1e6 + 5;
const int mod = 998244353;
int n, m, k, t, T, len, op, z, x, y;
int sum[N][27], a[N];
int check(int x, int y) {
  int l = 1, r = n, res = 0, t1 = 0, t2 = 0;
  res = max(res, sum[r][y] - sum[l - 1][y]);
  while (l <= n) {
    while (l <= n && a[l] != x) l++;
    while (r >= 1 && a[r] != x) r--;
    if (l > r) break;
    t1++;
    l++;
    if (l > r) break;
    res = max(res, sum[r][y] - sum[l - 1][y] + 2 * min(t1, t2));
    t2++;
    r--;
    if (l > r) break;
    res = max(res, sum[r][y] - sum[l - 1][y] + 2 * min(t1, t2));
  }
  return res;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 26; j++) sum[i][j] = sum[i - 1][j];
    scanf("%d", &a[i]);
    sum[i][a[i]]++;
  }
  int ans = 0;
  for (int i = 1; i <= 26; i++) {
    for (int j = 1; j <= 26; j++) {
      if (i == j) continue;
      ans = max(ans, check(i, j));
    }
  }
  printf("%d\n", ans);
}
int main() {
  int o;
  scanf("%d", &o);
  while (o--) {
    solve();
  }
  return 0;
}
