description: You are given n integers a_1, a_2, …, a_n. Find the maximum value of max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r) over all pairs (l, r) of integers for which 1 ≤ l < r ≤ n.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.
  
  The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
  
  Output
  
  For each test case, print a single integer — the maximum possible value of the product from the statement.
  
  Example
  
  Input
  
  
  4
  3
  2 4 3
  4
  3 2 3 1
  2
  69 69
  6
  719313 273225 402638 473783 804745 323328
  
  
  Output
  
  
  12
  6
  4761
  381274500335
  
  Note
  
  Let f(l, r) = max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r).
  
  In the first test case, 
  
    * f(1, 2) = max(a_1, a_2) ⋅ min(a_1, a_2) = max(2, 4) ⋅ min(2, 4) = 4 ⋅ 2 = 8. 
    * f(1, 3) = max(a_1, a_2, a_3) ⋅ min(a_1, a_2, a_3) = max(2, 4, 3) ⋅ min(2, 4, 3) = 4 ⋅ 2 = 8. 
    * f(2, 3) = max(a_2, a_3) ⋅ min(a_2, a_3) = max(4, 3) ⋅ min(4, 3) = 4 ⋅ 3 = 12. 
  
  
  
  So the maximum is f(2, 3) = 12.
  
  In the second test case, the maximum is f(1, 2) = f(1, 3) = f(2, 3) = 6.
solution: #include <bits/stdc++.h>
  using namespace std;
  long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
  long long ceile(long long &a, long long &b) { return (a / b) + ((a % b) != 0); }
  class Graph {
   public:
    long long V;
    vector<vector<long long>> adj;
    Graph(long long V) {
      this->V = V;
      adj.resize(V, vector<long long>(0));
    }
    void addEdge(long long u, long long v) { adj[u].push_back(v); }
    bool DFS(long long s, long long V) {
      vector<bool> vis(V, 0);
      stack<long long> stack;
      stack.push(s);
      long long ans = 0;
      while (!stack.empty()) {
        s = stack.top();
        stack.pop();
        if (!vis[s]) {
          ans++;
          vis[s] = true;
        }
        for (auto i = adj[s].begin(); i != adj[s].end(); ++i)
          if (!vis[*i]) stack.push(*i);
      }
      return ans == V;
    }
  };
  vector<long long> SieveOfEratosthenes(long long n) {
    vector<long long> ans;
    vector<bool> prime(n + 1, 1);
    for (long long p = 2; p * p <= n; p++)
      if (prime[p])
        for (long long i = p * p; i <= n; i += p) prime[i] = false;
    for (long long p = 2; p <= n; p++)
      if (prime[p]) ans.push_back(p);
    return ans;
  }
  int32_t main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<long long> arr(n);
      for (long long &c : arr) cin >> c;
      long long ans = arr[0] * arr[1];
      for (long long i = 1; i < n - 1; i++) {
        ans = max({ans, arr[i] * arr[i + 1], arr[i] * arr[i - 1]});
      }
      cout << ans << "\n";
    }
  }
  