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 mod = 1e9 + 7;
const int N = 1e5 + 4;
const long long MOD = 998244353;
long long POW(long long base, long long exponent) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % MOD;
    exponent = exponent >> 1;
    base = (base * base) % MOD;
  }
  return result;
}
long long InvFac(long long x) { return POW(x, MOD - 2); }
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, mx1, mx2, mx = 0;
    cin >> n;
    int a[n], occ[n][26];
    memset(occ, 0, sizeof(occ));
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < 26; i++) {
      for (int j = 0; j < n; j++) {
        if (a[j] == i + 1) {
          if (!j)
            occ[j][i] = 1;
          else
            occ[j][i] = occ[j - 1][i] + 1;
        } else if (j)
          occ[j][i] = occ[j - 1][i];
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = i; j < n; j++) {
        mx1 = 0, mx2 = 0;
        for (int k = 0; k < 26; k++) {
          if (i)
            mx1 = max(mx1, occ[j][k] - occ[i - 1][k]);
          else
            mx1 = max(mx1, occ[j][k]);
        }
        for (int k = 0; k < 26; k++) {
          if (i && j + 1 < n)
            mx2 = max(mx2, min(occ[i - 1][k], occ[n - 1][k] - occ[j][k]));
        }
        mx = max(mx, mx1 + 2 * mx2);
      }
    }
    cout << mx << '\n';
  }
  return 0;
}
