Problem: Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:

  * the following numbers are composite: 1024, 4, 6, 9; 
  * the following numbers are not composite: 13, 1, 2, 3, 37. 



You are given a positive integer n. Find two composite integers a,b such that a-b=n.

It can be proven that solution always exists.

Input

The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.

Output

Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).

It can be proven, that solution always exists.

If there are several possible solutions, you can print any. 

Examples

Input


1


Output


9 8


Input


512


Output


4608 4096

Solution (in PYTHON3):

Solution:
def check_compo(any_num):
	for i in range(2, int(any_num**0.5)+1):
		if any_num % i == 0:
			return True

	return False


num = int(input())
compo = 6

while True:
	b = num+compo
	if check_compo(b):
		print(b, compo)
		break
	else:
		compo += 1
		while check_compo(compo) is not True:
			compo += 1