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 int MxN = 2 * 1e3 + 3;
template <typename T>
void Write(T x) {
  if (x > 9) Write(x / 10);
  putchar(x % 10 + 48);
}
bool visit[MxN];
map<string, bool> mp;
bitset<MxN> bt;
void solve() {
  long long ans = 0, a[MxN], dp[MxN][27], n, hi[27], mx, vv[27], x = 0;
  memset(vv, 0, sizeof(vv));
  memset(hi, 0, sizeof(hi));
  scanf("%lld", &n);
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; ++i) scanf("%lld", &a[i]);
  for (int i = n - 1; ~i; --i) {
    hi[a[i]]++;
    x = max(x, hi[a[i]]);
    for (int j = 1; j <= 26; ++j) dp[i][j] = hi[j];
  }
  mx = 0;
  for (int i = 0; i < n; ++i) {
    vv[a[i]]++;
    long long v[27];
    memset(v, 0, sizeof(v));
    for (int j = i + 1; j < n; ++j) {
      v[a[j]]++;
      ans = max(ans, min(vv[a[i]], dp[j + 1][a[i]]) * 2 + v[a[j]]);
    }
  }
  cout << max(ans, x) << endl;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int q;
  scanf("%d", &q);
  while (q--) {
    solve();
  }
}
