Exactly even divisible using python recursion

def extractlyEvenlyDivisibleHelp(inputlst, number, outputlst = []):
    if number == 0:
        print([])
        return
    if inputlst[0] % number == 0:
        outputlst.append(inputlst[0])
        outputlst.sort()


    inputlst.remove(inputlst[0])

    if len(inputlst) == 0:
        print(outputlst)
    else:
        extractlyEvenlyDivisibleHelp(inputlst, number, outputlst)


def extractlyEvenlyDivisible(input, number):
    extractlyEvenlyDivisibleHelp(input, number, [])


extractlyEvenlyDivisible([1,2,3,4,5,6,7,8,9], 0)                                                
extractlyEvenlyDivisible([1,2,-3,4,5,-6,7,8,9,9,6], -3)
extractlyEvenlyDivisible([1,2,3,4,5,6,7,8,9,10, 10], 5)

Output:

[ ]

[-6, -3, 6, 9, 9]

[5, 10, 10]

Expected output:

[ ]

[-6, -3, 6, 9]

[5, 10]

I need help, as I need only one 9, I mean if I input couple times the same digit, it should return only one time.

You could filter out duplicates from the input list before passing to the helper using set:

def extractlyEvenlyDivisible(inputlst, number):
    extractlyEvenlyDivisibleHelp(list(set(inputlst)), number, [])

Also it is bad practice to redefine built-ins like input even in a function, I would change this to a different variable name