Solving Subset Product

import itertools
from itertools import combinations
from functools import reduce
from operator import mul
from sys import exit
# Given a whole number > 0 for TARGET, and a set of whole numbers > 0 that are divisors of TARGET. Is there a combination of divisors in SET that have a total product equivalent to TARGET?
TARGET = int(input('enter target product: ' ))
SET = input('Enter numbers: ').split(',')
SET = list(set([int(a) for a in SET]))
# Using two-sum solution for two-product
two_prod_dict = {}
for a in range(0, len(SET)):
    if TARGET//int(SET[a]) in two_prod_dict:
        print('YES')
        exit(0)
    else:
       two_prod_dict[int(SET[a])] = a
# creating a set of all whole number divisors
divisors = set()
for i in SET:
    if TARGET % i == 0:
        divisors.add(i)
# Divisors are sorted in ascending order.
# so that max_combo can correctly select
# the largest possible combination size.
divisors = sorted(divisors)
if TARGET in divisors:
    print('YES')
    exit(0)
# finds the maximum combination size
# Simply multiplies each I in divisors
# until reaches the closest
# value without going over.
# A combination's total product cannot be larger
# than the TARGET.
max_combo = []
for a in divisors:
    max_combo.append(a)
    if reduce(mul,max_combo) >= TARGET:
        if reduce(mul,max_combo) > TARGET:
            max_combo.pop()
            break
        else:
            break
# Using any other divisor will go over TARGET.
# To save running time, do it only one time.
for X in combinations(divisors, len(max_combo)):
    if reduce(mul, X) == TARGET:
        print('YES')
        exit(0)
    else:
        break
# I'm hoping that smaller solutions
# are more likely than larger solutions so start
# with the smallest and work your way up.
for A in range(3, len(max_combo) + 1):
    for X in combinations(divisors, A):
        if reduce(mul, X) == TARGET:
            print('YES',X)
            exit(0)
print('NO')