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>
#pragma GCC optimize("O3")
using namespace std;
long long int power(long long int x, 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;
}
long long int max3(long long int a, long long int b, long long int c) {
  return max(max(a, b), c);
}
long long int nCr(long long int n, long long int r) {
  long long int x = n - r, y = 0, z = 1;
  while (y < r) {
    z = (z * (x + 1)) / (y + 1);
    x++;
    y++;
  }
  return z;
}
long long int maxm = 1000000;
vector<long long int> v[201], v1;
long long int n, mx = -1;
void solve(long long int a, long long int b) {
  long long int i = 0, j = 0, sza = v[a].size(), szb = v[b].size();
  while (i < sza && j < szb) {
    if (v[a][i] < v[b][j]) {
      v1.push_back(a);
      i++;
    } else {
      v1.push_back(b);
      j++;
    }
  }
  while (i < sza) {
    v1.push_back(a);
    i++;
  }
  while (j < szb) {
    v1.push_back(b);
    j++;
  }
  i = 0, j = v1.size() - 1;
  long long int ca = 0, cb = szb;
  mx = max(mx, cb);
  while (i <= j) {
    if (v1[i] == b) {
      cb--;
      i++;
    } else if (v1[j] == b) {
      cb--;
      j--;
    } else {
      ca++;
      if (i == j) ca--;
      mx = max(mx, 2 * ca + cb);
      i++;
      j--;
    }
  }
  v1.clear();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long int i = 0; i < n; i++) {
      long long int x;
      cin >> x;
      v[x].push_back(i);
    }
    for (long long int i = 1; i < 201; i++)
      for (long long int j = 1; j < 201; j++)
        if (v[i].size() != 0 && v[j].size() != 0) solve(i, j);
    cout << mx << endl;
    for (long long int i = 0; i < 201; i++) v[i].clear();
    mx = -1;
  }
}
