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 MOD = 1E9 + 7;
  long long int gcd(long long int a, long long int b) {
    if (b == 0)
      return a;
    else
      return gcd(b, a % b);
  }
  long long int gcd_ex(long long int a, long long int b, long long int& x,
                       long long int& y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long int x1, y1;
    long long int g = gcd_ex(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1 * (a / b);
    return g;
  }
  long long int multi_mod_inv(long long int a, long long int m) {
    long long int x, y;
    long long int g = gcd_ex(a, m, x, y);
    if (g != 1) {
      return -1;
    }
    x = (x % m + m) % m;
    return x;
  }
  long long int mod_div(long long int a, long long int b, long long int m) {
    if (multi_mod_inv(b, m) != -1) {
      return (a % m * multi_mod_inv(b, m) % m) % m;
    } else {
      return -1;
    }
  }
  long long int lcm(long long int a, long long int b) {
    return a / gcd(a, b) * b;
  }
  long long int power(long long int a, long long int n) {
    if (n == 0) return 1;
    long long int p = power(a, n / 2);
    p = (p * p) % MOD;
    if (n % 2)
      return (p * a) % MOD;
    else
      return p % MOD;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    long long int test_case = 1;
    cin >> test_case;
    for (long long int test = 1; test <= test_case; test++) {
      long long int n;
      cin >> n;
      long long int ans = 0;
      vector<long long int> a(n);
      for (long long int i = 0; i < n; i++) {
        cin >> a[i];
      }
      for (long long int i = 1; i < n; i++) {
        ans = max(ans, a[i] * a[i - 1]);
      }
      cout << ans << '\n';
    }
    return 0;
  }
  