Count all subsequences whose product is less than K

def countSubsequences(a, n, p):
    dp={}
    '''
    We shall create a dp of size K*N k is given product.
    Each cell (x,y) in dp tells that how many subsequences are possible whose
    product <=k by choosing first y elements
    '''
    def help(i,cur):

        if cur==0 or i<0:return 0
        if (i,cur) in dp:return dp[(i,cur)]
        ans=0
        '''
        cur:current product
        i:first i elements of array a
        Now we are taking all the subsequence of previous cell
        since these are also the answers. And we shall add
        more subsequence if we found something new
        '''
        ans+=help(i-1,cur)

        if a[i]<=cur and a[i]>0:
            ans+=help(i-1,cur//a[i])+1
        '''
        We shall add 1 extra subsequence which is new {a[i]} .We shall add.
        Now we shall check the new subsequences with this a[i].
        so we shall compute (i-1,cur//a[i]). Here first parameter is i-1
        because we are computing it for first i vals of array.
        we shall count the subsequences of cur//a[i] that can be added to our a[i] and
        add this count to our answer
        '''
        dp[(i,cur)]=ans
        return ans
    return help(n-1,p)