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;
long long int power(long long int x, unsigned long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long int powermod(long long int x, unsigned long long int y,
                       long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
using namespace std;
long long int dp[27][2005];
long long int findFirstOccurrence(long long int arr[], long long int N,
                                  long long int x) {
  long long int low = 0, high = N - 1;
  long long int result = -1;
  while (low <= high) {
    long long int mid = (low + high) / 2;
    if (x == arr[mid]) {
      result = mid;
      high = mid - 1;
    } else if (x < arr[mid])
      high = mid - 1;
    else
      low = mid + 1;
  }
  return result;
}
long long int t, n, arr[2005];
bool f(long long int x) {
  for (long long int i = (long long int)(1); i < (long long int)(27); i++) {
    for (long long int j = (long long int)(1); j < (long long int)(n + 1);
         j++) {
      long long int ind =
          findFirstOccurrence(dp[i], n + 1, dp[i][n] - dp[i][j] + 1);
      if (ind > j) {
        for (long long int k = (long long int)(1); k < (long long int)(27);
             k++) {
          if (dp[k][ind - 1] - dp[k][j] + 2 * dp[i][j] >= x) return true;
        }
      }
    }
  }
  return false;
}
int32_t main() {
  cin >> t;
  while (t--) {
    memset(dp, 0, sizeof(dp));
    cin >> n;
    for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
      for (long long int j = (long long int)(0); j < (long long int)(27); j++) {
        dp[j][i + 1] = dp[j][i];
      }
      cin >> arr[i];
      dp[arr[i]][i + 1] += 1;
    }
    long long int low = 1, high = n, mid;
    while (low <= high) {
      mid = (high + low) / 2;
      if (f(mid)) {
        low = mid + 1;
      } else {
        high = mid - 1;
      }
    }
    cout << max(low - 1, 1ll) << endl;
    ;
  }
}
