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 int ksm(long long int a, long long int b, long long int mod) {
    long long int ans = 1;
    while (b) {
      if (b & 1) {
        ans = (a % mod * ans % mod) % mod;
      }
      b >>= 1;
      a = (a % mod * a % mod) % mod;
    }
    return ans;
  }
  long long int gcd(long long int a, long long int b) {
    return b == 0 ? a : gcd(b, a % b);
  }
  const long long int mod = 1e9 + 7;
  const int maxn = 2e5 + 10;
  const int maxm = 2e3 + 50;
  const double eps = 1e-8;
  const long long int inf = 0x3f3f3f3f;
  const long long int lnf = 0x3f3f3f3f3f3f3f3f;
  const double pi = acos(-1);
  int n, m;
  long long int a[maxn];
  string s;
  void sove() {
    long long int mi = inf;
    long long int ma = 0;
    cin >> n;
    long long int ans = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (i > 1) ans = max(ans, a[i] * a[i - 1]);
    }
    cout << ans << "\n";
  }
  int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    ;
    int T;
    cin >> T;
    while (T--) {
      sove();
    }
    return 0;
  }
  