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: import sys,os,time
  from math import log,sqrt,floor
  from os import path
  import math
  '''Input handler'''
  oj=True
  if path.exists('inp.txt'):
      print('YES')
      sys.stdin=open('inp.txt','r')
      sys.stdout=open('out.txt','w')
      oj=False
  '''Constants'''
  mod=1000000007
  INF=float('inf')
  
  
  '''Input functions'''
  def readint():
      return int(sys.stdin.readline())
  def readlst():
      return list(map(int,sys.stdin.readline().strip().split()))
  def readmul():
      return map(int,sys.stdin.readline().strip().split())
  def is_power_two(x):
      return ((x^(x-1)&x)==x)
  def lcm(a,b):
      return a*b//gcd(a, b)
  '''
  use for seive
  maxN=80000+10000+6
  primes=[True]*(maxN+1)
  '''
  def seive():
      primes[0],primes[1]=False,False
      for i in range(2,maxN+1):
          for j in range(i*i,maxN+1,i):
              if(primes[j]):
                  primes[j]=False
  def power_2(n):
      return 1<<n
  
  def ceil(a,b):
      return int((a+b-1)/b)
  
  def isPerfectSquare(x):
      if(x >= 0):
          sr = sqrt(x)
          return (sr*sr == x)
      return false
  
  
  
  def omkar():
      n=readint()
      l=readlst()
      mx=max(l)
      ans=0
      ind=[]
      for i in range(0,n):
          if(i-1>=0):
              ans=max(ans, l[i]*l[i-1])
          if(i+1<n):
              ans=max(ans, l[i]*l[i+1])
      print(ans)
      return
  
  def main():
      tc=readint()
      #tc=1
      start=time.time()
      while(tc):
          omkar()
          tc-=1
      if not oj:
          total=time.time()-start
          print(str(total*1000)+'ms')
  main()
  