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>
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using std::cin;
using std::cout;
using std::lower_bound;
using std::string;
using std::upper_bound;
using std::vector;
using vi = vector<long long>;
using vii = vector<vi>;
using pii = std::pair<long long, long long>;
constexpr long long MOD = 998244353;
constexpr long long MAX = 1e7;
constexpr long long INF = (1ll << 62);
template <class T>
class prique : public std::priority_queue<T, std::vector<T>, std::greater<T>> {
};
template <typename T>
struct Segment_tree {
  long long N;
  T mem;
  vector<T> node;
  Segment_tree(vector<T> &X, T m) : mem(m) {
    long long sz = X.size();
    N = 1;
    while (N < sz) N *= 2;
    node.resize(2 * N - 1, mem);
    for (int i = (int)(0); i < (int)(sz); i++) node[N - 1 + i] = X[i];
    for (int i = (int)(N - 2); (int)(0) <= i; i--) {
      node[i] = Compare(node[i * 2 + 1], node[i * 2 + 2]);
    }
  }
  T Compare(T &A, T &B) { return std::max(A, B); }
  void update(long long X, T val) {
    X += N - 1;
    node[X] = val;
    while (X > 0) {
      X = (X - 1) / 2;
      node[X] = Compare(node[X * 2 + 1], node[X * 2 + 2]);
    }
  }
  T Query(long long a, long long b, long long now, long long l, long long r) {
    if (r < 0) r = N;
    if (r <= a || b <= l) return mem;
    if (a <= l && r <= b) return node[now];
    auto vl = Query(a, b, now * 2 + 1, l, (l + r) / 2),
         vr = Query(a, b, now * 2 + 2, (l + r) / 2, r);
    return Compare(vl, vr);
  }
};
struct Binary_indexed_tree {
  int N;
  vi bit;
  Binary_indexed_tree(int n) : N(n) { bit.resize(N + 1, 0); }
  void add(int x, int a) {
    for (x; x <= N; x += (x & -x)) bit[x] += a;
  }
  long long sum(int x) {
    long long ret = 0;
    for (x; x > 0; x -= (x & -x)) ret += bit[x];
    return ret;
  }
  long long lower_bound(long long X) {
    if (sum(N) < X) return -1;
    long long ret = 0, memo = 1, sum = 0;
    while (memo * 2 <= N) memo *= 2;
    while (memo > 0) {
      if (memo + ret <= N && sum + bit[memo + ret] < X) {
        sum += bit[memo + ret];
        ret += memo;
      }
      memo /= 2;
    }
    return ret + 1;
  }
};
struct Union_Find {
  long long N;
  vi par;
  vi siz;
  Union_Find(int n) : N(n) {
    par.resize(N);
    siz.resize(N, 1);
    for (int i = (int)(0); i < (int)(N); i++) par[i] = i;
  }
  long long root(long long X) {
    if (par[X] == X) return X;
    return par[X] = root(par[X]);
  }
  bool same(long long X, long long Y) { return root(X) == root(Y); }
  void unite(long long X, long long Y) {
    X = root(X);
    Y = root(Y);
    if (X == Y) return;
    par[X] = Y;
    siz[Y] += siz[X];
    siz[X] = 0;
  }
  long long size(long long X) { return siz[root(X)]; }
};
long long modpow(long long a, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
vi fac, finv, inv;
void COMinit() {
  fac.resize(MAX);
  finv.resize(MAX);
  inv.resize(MAX);
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}
long long COM(long long n, long long r) {
  if (n < r || n < 0 || r < 0) return 0;
  return fac[n] * finv[r] % MOD * finv[n - r] % MOD;
}
long long modinv(long long a, long long m) {
  long long b = m, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b;
    std::swap(a, b);
    u -= t * v;
    std::swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  long long T;
  cin >> T;
  while (T--) {
    long long N;
    cin >> N;
    vi A(N);
    for (int i = (int)(0); i < (int)(N); i++) cin >> A[i];
    vii sum(N + 1, vi(27));
    for (int i = (int)(0); i < (int)(N); i++) {
      for (int j = (int)(0); j < (int)(27); j++) sum[i + 1][j] = sum[i][j];
      sum[i + 1][A[i]]++;
    }
    long long ans = 0;
    for (int i = (int)(1); i <= (int)(N); i++) {
      for (int j = (int)(0); j < (int)(i); j++) {
        long long memo = 0;
        for (int k = (int)(0); k < (int)(27); k++)
          memo = std::max(memo, sum[i][k] - sum[j][k]);
        for (int k = (int)(0); k < (int)(27); k++) {
          ans = std::max(ans,
                         std::min(sum[j][k], sum[N][k] - sum[i][k]) * 2 + memo);
        }
      }
    }
    cout << ans << "\n";
  }
}
