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>
const long long INF = 1LL << 60;
const long long MOD = 1000000007;
using namespace std;
using ll = long long;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
ll devc(ll x, ll y) { return (x + y - 1) / y; }
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(20);
  ll t;
  cin >> t;
  for (ll _ = 0; _ < (t); ++_) {
    ll n;
    cin >> n;
    vector<ll> a(n);
    for (ll i = 0; i < (n); ++i) cin >> a[i];
    vector<vector<ll>> f(n + 1, vector<ll>(201, 0LL));
    for (ll i = 0; i < (n); ++i) ++f[i + 1][a[i]];
    for (ll i = 0; i < (n); ++i)
      for (ll j = 0; j < (201); ++j) f[i + 1][j] += f[i][j];
    ll ans = 1;
    for (ll i = 1; i <= (n + 1); ++i) {
      if (i == 1) {
        for (ll j = 1; j <= (200); ++j) chmax(ans, f[n][j]);
        continue;
      }
      ll k = a[i - 2];
      ll ct = f[i - 1][k];
      ll l = -1, r = n + 1;
      while (r - l > 1) {
        ll m = (r + l) / 2;
        if (f[n][k] - f[m][k] < ct)
          r = m;
        else
          l = m;
      }
      if (l < i) continue;
      ll tp = 0;
      for (ll x = 1; x <= (200); ++x) chmax(tp, f[l][x] - f[i - 1][x]);
      chmax(ans, tp + 2 * ct);
    }
    cout << (ans) << '\n';
  }
  return 0;
}
