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;
  using namespace chrono;
  template <class T>
  inline bool chmax(T &a, T b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T>
  inline bool chmin(T &a, T b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  template <typename Arg1>
  void __f(const char *name, Arg1 &&arg1) {
    cerr << name << " : " << arg1 << std::endl;
  }
  template <typename Arg1, typename... Args>
  void __f(const char *names, Arg1 &&arg1, Args &&...args) {
    const char *comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cerr << '{';
    for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  const long long int inf = 2e18;
  const long long int mod = 1e9 + 7;
  const long long int N = 2e5 + 3;
  int main() {
    auto start_time = high_resolution_clock::now();
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int tc;
    cin >> tc;
    while (tc--) {
      long long int n;
      cin >> n;
      vector<long long int> v(n);
      for (auto i = 0; i < n; ++i) cin >> v[i];
      long long int ans = 0;
      for (auto i = 0; i < n - 1; ++i) {
        chmax(ans, v[i] * v[i + 1]);
      }
      cout << ans << "\n";
    }
    auto end_time = high_resolution_clock::now();
    return 0;
    cout << "\nTime elapsed: "
         << (double)duration_cast<milliseconds>(end_time - start_time).count()
         << "ms";
    return 0;
  }
  