findTwoInitialNumbers

Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y(including 1 and y). If z is a divisor of both numbers x and y at the same time, there are two occurrences of z in the list.

For example, if x=4 and y=6 then the given list can be any permutation of the list [1,2,4,1,2,3,6]. Some of the possible lists are: [1,1,2,4,6,3,2], [4,6,1,1,2,3,2] or [1,6,3,2,4,1,2].

Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).

It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.

Example:

• For n = 10, listOfDivisors = [10, 2, 8, 1, 2, 4, 1, 20, 4, 5].
  The output should be:findTwoInitialNumbers=[20,8].
  Because:
  Numbers that are divisible by 20: 1,2,4,5,10,20.
  Numbers that are divisible by 8: 1,2,4,8.
  When we shuffle these numbers into a list. We has a list contains all divisors of these 2 numbers.
  Note that the given listOfDivisors is one of permutations of the above list.

Input/Output

• [Execution time limit] 0.5s in C++, 3s in Java and C#, 4s in Python, Go and JS
• [Input] interger n
  the number of divisors of x and y.
  2 ≤ n ≤ 128
• [Input] array.interger listOfDivisors
  The list contains all divisors of either x and y. If there is a number z such that z is divisible by both x and y, z will appear 2 times in this list.
  1 ≤ listOfDivisors[i] ≤ 104
• [Output] array.interger
  The initial numbers x and y
  Note that x >= y