Prime in Numbers with Wheel Factorization

Given a positive number n > 1 find the prime factor decomposition of n. The result will be a string with the following form :

"(p1**n1)(p2**n2)...(pk**nk)"
with the p(i) in increasing order and n(i) empty if n(i) is 1.
@EXAMPLE:
n = 86240 should return "(2**5)(5)(7**2)(11)"
n= 782611830 should return "(2)(3**2)(5)(7**2)(11)(13)(17)(73)"


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import math

def prime_factors(n):
    result=''
    count = 0
    while n%2==0:
        n/=2
        count+=1
    if count==1:
        x2=("("+"2"+")")
        for cnt01 in x2:
            result+=cnt01
    elif count:
        x= ("(" + "2**" + str(count) + ")" )
        for cnt02 in x:
            result+=cnt02


    for k in range(3,int(math.sqrt(n)+1),2):
        count=0
        while n%k==0 and n!=1:
            n/=k
            count+=1
        if count==1:
            z=("("+str(k)+")")
            for cnt in z:
                result+=cnt

        elif count:
            y= ("("+str(k) + "**" + str(count)+")")
            for cnt1 in y:
                result+=cnt1
    if n > 2:
        x1=("("+str(int(n))+")")
        for cnt2 in x1:
            result+=cnt2

    return result