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,io
  # input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  PI = 3.141592653589793238460
  INF =  float('inf')
  MOD  = 1000000007
  # MOD = 998244353
  
  
  def bin32(num):
      return '{0:032b}'.format(num)
  
  def add(x,y):
      return (x+y)%MOD
  
  def sub(x,y):
      return (x-y+MOD)%MOD
  
  def mul(x,y):
      return (x*y)%MOD
  
  def gcd(x,y):
      if y == 0:
          return x
      return gcd(y,x%y)
  
  def lcm(x,y):
      return (x*y)//gcd(x,y)
  
  def power(x,y):
      res = 1
      x%=MOD
      while y!=0:
          if y&1 :
              res = mul(res,x)
          y>>=1
          x = mul(x,x)
          
      return res
          
  def mod_inv(n):
      return power(n,MOD-2)
  
  def prob(p,q):
      return mul(p,power(q,MOD-2))    
    
  def ii():
      return int(input())
  
  def li():
      return [int(i) for i in input().split()]
  
  def ls():
      return [i for i in input().split()]
  
  for t in range(ii()):
      t+=1
      n = ii()
      a= li()
      ans = 0
      for i in range(n-1):
          ans = max(ans , a[i] * a[i+1])
      print(ans)