Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number n. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that n is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer x was given. The task was to add x to the sum of the digits of the number x written in decimal numeral system.

Since the number n on the board was small, Vova quickly guessed which x could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number n for all suitable values of x or determine that such x does not exist. Write such a program for Vova.

Input

The first line contains integer n (1 ≤ n ≤ 109).

Output

In the first line print one integer k — number of different values of x satisfying the condition. 

In next k lines print these values in ascending order.

Examples

Input

21


Output

1
15


Input

20


Output

0

Note

In the first test case x = 15 there is only one variant: 15 + 1 + 5 = 21.

In the second test case there are no such x.
#winners never quit, quiters never win
from collections import deque as de
import math
from collections import Counter as cnt
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0

arrStack = My_stack()    
# A optimized school method based 
# Python3 program to check 
# if a number is prime 


def isPrime(n) : 

	# Corner cases 
	if (n <= 1) : 
		return False
	if (n <= 3) : 
		return True

	# This is checked so that we can skip 
	# middle five numbers in below loop 
	if (n % 2 == 0 or n % 3 == 0) : 
		return False

	i = 5
	while(i * i <= n) : 
		if (n % i == 0 or n % (i + 2) == 0) : 
			return False
		i = i + 6

	return True

def get_prime_factors(number):
    # create an empty list and later I will
    # run a for loop with range() function using the append() method to add elements to the list.
    prime_factors = []

    # First get the number of two's that divide number
    # i.e the number of 2's that are in the factors
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2

    # After the above while loop, when number has been
    # divided by all the 2's - so the number must be odd at this point
    # Otherwise it would be perfectly divisible by 2 another time
    # so now that its odd I can skip 2 ( i = i + 2) for each increment
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i


    # Here is the crucial part.
    # First quick refreshment on the two key mathematical conjectures of Prime factorization of any non-Prime number
    # Which is - 1. If n is not a prime number AT-LEAST one Prime factor would be less than sqrt(n)
    # And - 2. If n is not a prime number - There can be AT-MOST 1 prime factor of n greater than sqrt(n).
    # Like 7 is a prime-factor for 14 which is greater than sqrt(14)
    # But if the above loop DOES NOT go beyond square root of the initial n.
    # Then how does that greater than sqrt(n) prime-factor
    # will be captured in my prime factorization function.
    # ANS to that is - in my first for-loop I am dividing n with the prime number if that prime is a factor of n.
    # Meaning, after this first for-loop gets executed completely, the adjusted initial n should become
    # either 1 or greater than 1
    # And if n has NOT become 1 after the previous for-loop, that means that
    # The remaining n is that prime factor which is greater that the square root of initial n.
    # And that's why in the next part of my algorithm, I need to check whether n becomes 1 or not,
    #This code is taken by rohan paul's github
    if number > 2:
        prime_factors.append(int(number))

    return prime_factors
def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic
#here we go......................
#winners never quit, quitters never win
n=int(input())
temp=n
temp=str(temp)
ans=[]
if n <=10 :
    if n%2:
        print(0)
    else:
        print(1)
        print(n//2)
else:
    ll=((len(temp)-1)*9)+ (int(temp[0])-1)
    for i in range(1, ll+1):
        dd=str(n-i)
        if i== sum(list(map(int, dd.strip()))):
            ans.append(int(dd))
    if len(ans)==0:
        print(0)
    else:
        print(len(ans))
        ans.sort()
        print(*ans, end="\n")
            
        
        


