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;
int power(int x, unsigned int y) {
  int res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int powermod(int x, unsigned int y, int p) {
  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;
int32_t main() {
  int t, n, arr[100005], fr, ba, cf, cb, ele, freq, ans;
  cin >> t;
  while (t--) {
    cin >> n;
    ans = 1;
    for (int i = (int)(0); i < (int)(n); i++) {
      cin >> arr[i];
    }
    for (int i = (int)(1); i < (int)(201); i++) {
      cf = 0;
      cb = 0;
      set<pair<int, int> > ml;
      map<int, int> mp;
      fr = 0;
      ba = n - 1;
      for (int j = (int)(0); j < (int)(n); j++) {
        mp[arr[j]] += 1;
      }
      for (auto &val : mp) {
        ans = max(ans, val.second);
        ml.insert({-val.second, val.first});
      }
      while (fr <= ba) {
        while (fr <= ba and arr[fr] != i) {
          freq = mp[arr[fr]];
          ml.erase({-freq, arr[fr]});
          mp[arr[fr]] -= 1;
          freq -= 1;
          ml.insert({-freq, arr[fr]});
          fr += 1;
        }
        if (fr <= ba) {
          freq = mp[arr[fr]];
          ml.erase({-freq, arr[fr]});
          mp[arr[fr]] -= 1;
          freq -= 1;
          ml.insert({-freq, arr[fr]});
          fr += 1;
          cf += 1;
        }
        while (fr <= ba and arr[ba] != i) {
          freq = mp[arr[ba]];
          ml.erase({-freq, arr[ba]});
          mp[arr[ba]] -= 1;
          freq -= 1;
          ml.insert({-freq, arr[ba]});
          ba -= 1;
        }
        if (fr <= ba) {
          freq = mp[arr[ba]];
          ml.erase({-freq, arr[ba]});
          mp[arr[ba]] -= 1;
          freq -= 1;
          ml.insert({-freq, arr[ba]});
          ba -= 1;
          cb += 1;
          ans = max(ans, cf + cb - ml.begin()->first);
        }
      }
    }
    cout << ans << endl;
  }
}
