Let S(n) denote the sum of the digits in the decimal notation of n. For example, S(123) = 1 + 2 + 3 = 6.

We will call an integer n a Snuke number when, for all positive integers m such that m > n, \frac{n}{S(n)} \leq \frac{m}{S(m)} holds.

Given an integer K, list the K smallest Snuke numbers.

Constraints

* 1 \leq K
* The K-th smallest Snuke number is not greater than 10^{15}.

Input

Input is given from Standard Input in the following format:


K


Output

Print K lines. The i-th line should contain the i-th smallest Snuke number.

Example

Input

10


Output

1
2
3
4
5
6
7
8
9
19
import math
# import datetime

def s(n):
	sum = 0
	for c in str(n):
		sum += int(c)
	return sum

k = int(input())
found=0

def next_sunuke(n, s_n, s_n_div):

	global found

	if n == 1:
		return

	log_n = int(math.log(n,10))
	m = n - 10 ** int(log_n // 1.25)

	while m >= 1:
		s_m = s(m)
		s_mplus1 = s(m + 1)

		if (s_m < s_mplus1 or (s_n - s_m != 1 and s_n - s_m != -8 and s_n - s_m != -17)) and m > 9:
			m -= 10 ** int(log_n // 1.25)
			continue

		# print("check: " + str(m))

		s_m_div = m / s(m)

		if s_m_div <= s_n_div:
			next_sunuke(m, s_m, s_m_div)
			if found >= k:
				return

			found += 1
			# print("{0:d}".format(m), "{0:+d}".format(n-m), "{0:d}".format(s_m), s_m_div)
			print(m)
			break


		m -= 10 ** int(log_n // 1.25)


# 適当な大きいスヌケ数
current_sunuke = 999999999999999
s_current_sunuke = s(current_sunuke)
s_current_sunuke_div = current_sunuke / s_current_sunuke

# start = datetime.datetime.now()
next_sunuke(current_sunuke, s_current_sunuke, s_current_sunuke_div)
if found < k:
	print(current_sunuke)
# end = datetime.datetime.now()
# print(str(end-start))
