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;
struct tree {
  int f[205];
};
void build(struct tree t[], long long n) {
  for (long long i = n - 1; i > 0; i--) {
    for (long long j = 0; j < 205; j++) {
      t[i].f[j] = t[i << 1].f[j] + t[i << 1 | 1].f[j];
    }
  }
}
int query(struct tree t[], long long n, long long l, long long r) {
  int res[205];
  for (long long i = 0; i < 205; i++) res[i] = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      for (long long j = 0; j < 205; j++) res[j] += t[l].f[j];
      l++;
    }
    if (r & 1) {
      --r;
      for (long long j = 0; j < 205; j++) res[j] += t[r].f[j];
    }
  }
  int ans = 0;
  for (long long j = 0; j < 205; j++) {
    ans = max(ans, res[j]);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int a[n];
    struct tree t[2 * n];
    int f[205] = {0};
    int ans = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      f[a[i]]++;
      ans = max(ans, f[a[i]]);
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 205; j++) {
        t[n + i].f[j] = 0;
      }
      t[n + i].f[a[i]]++;
    }
    build(t, n);
    for (int i = 0; i < 205; i++) {
      if (f[i] <= 1) continue;
      int x = f[i] / 2, cnt = 0;
      int l = 0, r = n - 1;
      vector<long long> L, R, W;
      while (l < n) {
        if (a[l] == i) cnt++;
        l++;
        if (a[l - 1] == i) {
          L.push_back(l);
          W.push_back(cnt);
        }
        if (cnt == x) break;
      }
      while (r >= 0) {
        if (a[r] == i) cnt--;
        r--;
        if (a[r + 1] == i) R.push_back(r);
        if (cnt == 0) break;
      }
      int tmp = 0, mx = 0;
      for (int j = 0; j < L.size(); j++) {
        mx = query(t, n, L[j], R[j] + 1);
        tmp = max(tmp, (int)(2 * W[j] + mx));
      }
      ans = max(ans, tmp);
    }
    cout << ans << '\n';
  }
}
