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>
  #pragma GCC optimize(2)
  using namespace std;
  const int inf = 0x3f3f3f3f;
  const int MOD = 1e9 + 7;
  const double eps = 1e-6;
  const double pi = acos(-1.0);
  void debug() { cerr << '\n'; }
  template <typename T, typename... Ts>
  void debug(T x, Ts... y) {
    cerr << "\033[31m" << x << "\033[0m";
    if (sizeof...(y) > 0) cerr << ' ';
    debug(y...);
  }
  template <typename T>
  void debug(const T &a, int l, int r, char c) {
    for (int i = l; i <= r; ++i)
      cerr << "\033[31m" << a[i] << "\033[0m" << (i == r ? '\n' : c);
  }
  template <typename T>
  void debug(vector<T> a) {
    for (int i = 0; i < (int)a.size(); ++i)
      cerr << "\033[31m" << a[i] << "\033[31m"
           << " \n"[i == ((int)a.size() - 1)];
  }
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  long long qpow(long long x, long long y) {
    long long ans = 1;
    for (; y > 0; y >>= 1) {
      if (y & 1) ans *= x;
      x *= x;
    }
    return ans;
  }
  long long qpow(long long x, long long y, int MOD) {
    long long ans = 1;
    for (; y > 0; y >>= 1) {
      if (y & 1) ans = ans * x % MOD;
      x = x * x % MOD;
    }
    return ans;
  }
  void exgcd(long long a, long long b, long long &x, long long &y) {
    if (!b) {
      x = 1;
      y = 0;
      return;
    }
    exgcd(b, a % b, x, y);
    long long X = x;
    x = y;
    y = X - a / b * y;
  }
  long long inv(long long x, int mod = MOD) { return qpow(x, mod - 2, mod); }
  long long m_m(long long a) { return (a % MOD + MOD) % MOD; }
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
      int n;
      cin >> n;
      vector<int> a(n);
      for (auto &it : a) cin >> it;
      long long ma = 0;
      for (int i = (1); i < (n); ++i) ma = max(ma, 1ll * a[i] * a[i - 1]);
      cout << ma << '\n';
    }
    return 0;
  }
  