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: // Working program with FastReader 
  import java.io.*;
  import java.util.*;
  public class ll
  { 
      static class FastReader 
      { 
          BufferedReader br; 
          StringTokenizer st; 
  
          public FastReader() 
          { 
              br = new BufferedReader(new
                  InputStreamReader(System.in)); 
          } 
  
          String next() 
          { 
              while (st == null || !st.hasMoreElements()) 
              { 
                  try
                  { 
                      st = new StringTokenizer(br.readLine()); 
                  } 
                  catch (IOException e) 
                  { 
                      e.printStackTrace(); 
                  } 
              } 
              return st.nextToken(); 
          } 
  
          int nextInt() 
          { 
              return Integer.parseInt(next()); 
          } 
  
          long nextLong() 
          { 
              return Long.parseLong(next()); 
          } 
  
          double nextDouble() 
          { 
              return Double.parseDouble(next()); 
          } 
  
          String nextLine() 
          { 
              String str = ""; 
              try
              { 
                  str = br.readLine(); 
              } 
              catch (IOException e) 
              { 
                  e.printStackTrace(); 
              } 
              return str; 
          } 
      } 
  
      static final int MAXN = 100001;
      // stores smallest prime factor for every number
      static int spf[] = new int[MAXN];
      // Calculating SPF (Smallest Prime Factor) for every
      // number till MAXN.
      // Time Complexity : O(nloglogn)
      static void sieve()
      {
          spf[1] = 1;
          for (int i=2; i<MAXN; i++)
  
          // marking smallest prime factor for every
          // number to be itself.
              spf[i] = i;
  
          // separately marking spf for every even
          // number as 2
          for (int i=4; i<MAXN; i+=2)
              spf[i] = 2;
  
          for (int i=3; i*i<MAXN; i++)
          {
              // checking if i is prime
              if (spf[i] == i)
              {
                  // marking SPF for all numbers divisible by i
                  for (int j=i*i; j<MAXN; j+=i)
  
                  // marking spf[j] if it is not
                  // previously marked
                      if (spf[j]==j)
                          spf[j] = i;
              }
          }
      }
      // A O(log n) function returning primefactorization
      // by dividing by smallest prime factor at every step
      static int getFactorization(int x)
      {
          int c=0;
          while (x != 1)
          {
              c++;
              x = x / spf[x];
          }
          return c;
      }
  
      /*
      static int LowerBound(int a[], int x) { // x is the target value or key
      int l=-1,r=a.length;
      while(l+1<r) {
      int m=(l+r)>>>1;
      if(a[m]>=x) r=m;
      else l=m;
      }
      return r;
      }
  
      static int UpperBound(int a[], int x) {// x is the key or target value
      int l=-1,r=a.length;
      while(l+1<r) {
      int m=(l+r)>>>1;
      if(a[m]<=x) l=m;
      else r=m;
      }
      return l+1;
      }
       */
      static int gcd(int a, int b)
      {
          if (a == 0)
              return b; 
          return gcd(b % a, a); 
      }
  
      // method to return LCM of two numbers
      static int lcm(int a, int b)
      {
          return (a / gcd(a, b)) * b;
      }
  
      public static int[] swap(int a[], int left, int right)
      {
          int temp = a[left];
          a[left] = a[right];
          a[right] = temp;
          return a;
      }
  
      public static int[] reverse(int a[], int left, int right)
      {
          // Reverse the sub-array
          while (left < right) {
              int temp = a[left];
              a[left++] = a[right];
              a[right--] = temp;
          }
          return a;
      }
  
      public static int[] findNextPermutation(int a[])
      {
          int last = a.length - 2;
  
          // find the longest non-increasing suffix
          // and find the pivot
          while (last >= 0) {
              if (a[last] < a[last + 1]) {
                  break;
              }
              last--;
          }
          // If there is no increasing pair
          // there is no higher order permutation
          if (last < 0)
              return a;
  
          int nextGreater = a.length - 1;
  
          // Find the rightmost successor to the pivot
          for (int i = a.length - 1; i > last; i--) {
              if (a[i] > a[last]) {
                  nextGreater = i;
                  break;
              }
          }
  
          // Swap the successor and the pivot
          a = swap(a, nextGreater, last);
  
          // Reverse the suffix
          a = reverse(a, last + 1, a.length - 1);
  
          // Return true as the next_permutation is done
          return a;
      }
  
      static void sort(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a)
              l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      static void sort(char[] a) {
          ArrayList<Character> l = new ArrayList<>();
          for (char i : a)
              l.add(i);
          Collections.sort(l);
          for (char i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      static double pow(double p,double tt)
      {
          double ii,q,r;
          q=1;
          r=p;
          while(tt>1)
          {
              for(ii=1;2*ii<=tt;ii*=2)
                  p*=p;
              tt-=ii;
              q*=p;
              p=r;
          }
          if(tt==1)
              q*=r;
          return q;
      }
  
      static long pow(long p,long tt,long mod)
      {
          long ii,q,r;
          q=1l;
          r=p;
          while(tt>1)
          {
              for(ii=1l;2*ii<=tt;ii*=2l)
                  p=((p%mod)*(p%mod))%mod;
              tt-=ii;
              q=((q%mod)*(p%mod))%mod;
              p=r;
          }
          if(tt==1)
              q=((q%mod)*(r%mod))%mod;
          return q;
      }
  
      static int factorial(int n)
      {
          return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
      }
  
      public static long primeFactors(long n)
      {
          long c=0l;
          long max=0l;
          long z=0l;
          //ArrayList <Integer> ll=new ArrayList<>();
          // Print the number of 2s that divide n
          while (n%2==0)
          {
              c++;
              n /= 2l;
              //ll.add(2);
          }
          if(c>max)
          {
              max=c;
              z=2;
          }
          // n must be odd at this point. So we can
          // skip one element (Note i = i +2)
          for (int i = 3; i <= Math.sqrt(n); i+= 2)
          {
              // While i divides n, print i and divide n
              c=0;
              while (n%i == 0)
              {
                  c++;
                  n /= i;
                  //ll.add(i);
              }
              if(c>max)
              {
                  max=c;
                  z=i;
              }
          }
          c=0;
          // This condition is to handle the case whien
          // n is a prime number greater than 2
          if (n > 2)
          {
              c++;
              //ll.add((int)n);
          }
          if(c>max)
          {
              max=c;
              z=n;
          }
          return z;
      }
  
      static void PrimeList(){
          int i,j;
          int sieve[]=new int[100001];
          for(i=2;i*i<=100000;i++)
          {
              if(sieve[i]==0)
              {
                  for(j=i*i;j<=100000;j+=i)
                      sieve[j]=1;
              }
          }
          ArrayList<Integer> primes=new ArrayList<>();
          for(i=2;i<=100000;i++)
          {
              if(sieve[i]==0)
                  primes.add(i);
          }
      }
  
      static int comp(int a[],int b[],int n)
      {
          int z=0;
          for(int i=0;i<n;i++)
          {
              if(a[i]!=b[i])
              {
                  z=1;
                  break;
              }
          }
          if(z==0)
              return 1;
          else
              return 0;
      }
  
      static boolean isPowerOfTwo(int x)
      {
          return x!=0 && ((x&(x-1))==0);
      }
  
      public static void main(String[] args) 
      {
          //try
          //{
          FastReader d=new FastReader();
          PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
          int t,i,j,k,l,r,n;
          int mod = (int) 1e9 + 7;
  
          int Inf=Integer.MAX_VALUE;
          int negInf=Integer.MIN_VALUE;
          t=d.nextInt();
          //t=1;
          String s;
          //char ch1,ch2,ch3,ch4;
          long ans,c,z;
  
          while(t-->0)
          {
              z=c=0l;
              ans=0l;
              n=d.nextInt();
              long a[]=new long[n];
              for(i=0;i<n;i++)
              a[i]=d.nextLong();
              long max=0l;
              for(i=0;i<n-1;i++)
              {
                  if(a[i]*a[i+1]>max)
                  max=(long)a[i]*(long)a[i+1];
              }
              System.out.println(max);
              //pr.println(ans);
          }
          /*
          }catch(Exception e) {
          System.out.println(0);
          }*/  
          pr.flush();
      }
  }