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;
template <class T>
void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void chmin(T &a, T b) {
  if (a > b) a = b;
}
constexpr long long INF = 1000000000000000000;
constexpr long long mod = 1000000007;
long long dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
long long kaijo[2000010];
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
bool prime(long long a) {
  if (a == 1) return false;
  for (long long i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
void init_fact() {
  kaijo[0] = 1;
  for (long long i = 1; i <= 2000000; i++) {
    kaijo[i] = kaijo[i - 1] * i;
    kaijo[i] %= mod;
  }
}
long long modpow(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2) return modpow(a, b - 1) * a % mod;
  long long memo = modpow(a, b / 2);
  return memo * memo % mod;
}
long long comb(long long a, long long b) {
  if (!kaijo[0]) init_fact();
  return kaijo[a] * modpow(kaijo[a - b], mod - 2) % mod *
         modpow(kaijo[b], mod - 2) % mod;
}
long long inv(long long x) { return modpow(x, mod - 2); }
bool kosa(double ax, double ay, double bx, double by, double cx, double cy,
          double dx, double dy) {
  double ta = (cx - dx) * (ay - cy) + (cy - dy) * (cx - ax);
  double tb = (cx - dx) * (by - cy) + (cy - dy) * (cx - bx);
  double tc = (ax - bx) * (cy - ay) + (ay - by) * (ax - cx);
  double td = (ax - bx) * (dy - ay) + (ay - by) * (ax - dx);
  return tc * td < 0 && ta * tb < 0;
}
long long n, a[200010];
void solve() {
  cin >> n;
  vector<long long> pos[200];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
    pos[a[i]].push_back(i);
  }
  long long ans = 0;
  for (long long c = 0; c < 200; c++) {
    if (pos[c].empty()) continue;
    for (long long i = 0; i * 2 <= pos[c].size(); i++) {
      long long l, r;
      if (i == 0)
        l = 0, r = n - 1;
      else
        l = pos[c][i - 1] + 1, r = pos[c][pos[c].size() - i] - 1;
      long long ma = 0;
      for (long long j = 0; j < 200; j++) {
        long long cnt = upper_bound(pos[j].begin(), pos[j].end(), r) -
                        lower_bound(pos[j].begin(), pos[j].end(), l);
        chmax(ma, cnt);
      }
      chmax(ans, ma + i * 2);
    }
  }
  cout << ans << endl;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}
