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 power(long long int a, long long int n) {
    if (n == 0)
      return 1;
    else if (n == 1)
      return a;
    else {
      if (n % 2 == 0)
        return (power(a, n / 2) % 1000000007 * power(a, n / 2) % 1000000007) %
               1000000007;
      else
        return (power(a, n / 2) % 1000000007 * power(a, n / 2) % 1000000007 * a) %
               1000000007;
    }
  }
  bool poweroftwo(int x) { return x && (!(x & (x - 1))); }
  long long int cmpi(pair<long long int, long long int> a,
                     pair<long long int, long long int> b) {
    return a.first < b.first;
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long int search(long long int a[], long long int n, long long int x) {
    long long int l = 0, r = n - 1;
    if (x < a[l] || x > a[r]) return -1;
    while (l <= r) {
      long long int mid = l + (r - l) / 2;
      if (a[mid] == x) return 1;
      if (a[mid] < x)
        l = mid + 1;
      else
        r = mid - 1;
    }
    return -1;
  }
  long long int isprime(long long int x) {
    long long int flag = 0;
    for (long long int i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        flag++;
        break;
      }
    }
    if (flag) return 1;
    return 0;
  }
  void solve() {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long int ans = LLONG_MIN;
    for (long long int i = 1; i < n - 1; i++) {
      ans = max(ans, max(min(a[i], a[i - 1]) * max(a[i], a[i - 1]),
                         min(a[i], a[i + 1]) * max(a[i], a[i + 1])));
    }
    ans = max(ans, a[0] * a[1]);
    ans = max(ans, a[n - 1] * a[n - 2]);
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    while (t--) solve();
  }
  