Two little greedy bears have found two pieces of cheese in the forest of weight *a* and *b* grams, correspondingly. The bears are so greedy that they are ready to fight for the larger piece. That's where the fox comes in and starts the dialog: "Little bears, wait a little, I want to make your pieces equal" "Come off it fox, how are you going to do that?", the curious bears asked. "It's easy", said the fox. "If the mass of a certain piece is divisible by two, then I can eat exactly a half of the piece. If the mass of a certain piece is divisible by three, then I can eat exactly two-thirds, and if the mass is divisible by five, then I can eat four-fifths. I'll eat a little here and there and make the pieces equal". 

The little bears realize that the fox's proposal contains a catch. But at the same time they realize that they can not make the two pieces equal themselves. So they agreed to her proposal, but on one condition: the fox should make the pieces equal as quickly as possible. Find the minimum number of operations the fox needs to make pieces equal.
The first line contains two space-separated integers *a* and *b* (1 ≤ *a*, *b* ≤ 10<sup class="upper-index">9</sup>).
from math import sqrt
 
# function to find the k'th divisor
def findkth(n, k):
     
    # initialize a vector v
    v = []
 
    # store all the divisors so the loop
    # will needs to run till sqrt ( n )
    p = int(sqrt(n)) + 1
    for i in range(1, p, 1):
        if (n % i == 0):
            v.append(i)
            if (i != sqrt(n)):
                v.append(n / i);
         
    # sort the vector in an increasing order
    v.sort(reverse = False)
 
    # if k is greater than the size of vector
    # then no divisor can be possible
    if (k > len(v)):
        print("Doesn't Exist")
         
    # else print the (k - 1)th
    # value of vector
    else:
        print(v[k - 1])
 
# Driver code
if __name__ == '__main__':
    n = 15
    k = 2
 
    findkth(n, k)
       	  	 	 	   	 		 			 			