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:
from math import sqrt
def mi():return map(int,input().split())
def li():return list(mi())
def ii():return int(input())
def si():return input()
mod=1000000007
def divisor(n):
    for i in range(2,int(sqrt(n))+1):
        if(n%i==0):
            return True
    return False
n=ii()
for i in range(2,1000000000):
    if(divisor(i) and divisor(i+n)):
        print(i+n,end=" ")
        print(i)
        break