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:
import sys
from math import sqrt, floor, factorial
from collections import deque, Counter
inp = sys.stdin.readline
read = lambda: list(map(int, inp().strip().split()))

def c_p(n):
	for i in range(2,int(sqrt(n))+1):
		if n % i == 0:
			return(True)
	return(False)
def solve():
	n = int(inp()); s = 4
	while not (c_p(s) and c_p(s+n)):
		s += 1
	print(s+n, s)


if __name__ == "__main__":
	solve()