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;
  inline void RI(long long &i) { scanf("%lld", &(i)); }
  inline void RVI(vector<long long> &v) {
    for (long long i = 0; i < long long((v).size()); ++i) {
      RI(v[i]);
    }
  }
  inline void RVVI(vector<vector<long long> > &vv) {
    for (long long i = 0; i < long long((vv).size()); ++i) {
      RVI(vv[i]);
    }
  }
  inline void WI(const long long &i) { printf("%lld\n", i); }
  inline void WVI(const vector<long long> &v, char sep = ' ') {
    for (long long i = 0; i < long long((v).size()); ++i) {
      if (i != 0) {
        printf("%c", sep);
      }
      printf("%lld", v[i]);
    }
    printf("\n");
  }
  inline void WS(const string &s) { printf("%s\n", s.c_str()); }
  inline void WB(bool b, const string &yes, const string &no) {
    if (b) {
      WS(yes);
    } else {
      WS(no);
    }
  }
  inline void YESNO(bool b) { WB(b, "YES", "NO"); }
  inline void RS(string &s) {
    static char buf[1000000];
    scanf("%s", buf);
    s = buf;
  }
  template <typename T>
  inline bool IN(T &S, const typename T::key_type &key) {
    return S.find(key) != S.end();
  }
  int main(int argc, char *argv[]) {
    long long i, j, k;
    long long t, T;
    RI(T);
    for (t = (0); t < (T); ++t) {
      long long n;
      RI(n);
      vector<long long> a(n);
      RVI(a);
      long long mv = *max_element((a).begin(), (a).end());
      long long ans = -1;
      for (i = (0); i < (n - 1); ++i) {
        ans = max(ans, a[i] * a[i + 1]);
      }
      WI(ans);
    }
    return 0;
  }
  