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 gcd
from heapq import heappop
from heapq import heappush
from heapq import heapify
from bisect import insort
from bisect import bisect_right
from bisect import bisect_left
from sys import stdin,stdout
from collections import defaultdict, deque
from math import log2, ceil, floor

inp=lambda : int(input())
sip=lambda : input()
mulip =lambda : map(int,input().split())
lst=lambda : list(map(int,input().split()))
slst=lambda: list(sip())
arr2d= lambda x: [[int(j) for j in input().split()] for i in range(x)]
odds = lambda l: len(list(filter(lambda x: x%2!=0, l)))
evens = lambda l: len(list(filter(lambda x: x%2==0, l)))
mod = pow(10,9)+7
#-------------------------------------------------------
Judge = 0
if Judge:
    sys.stdin = open("input.in",'r')
    #sys.stdout = open("output.in",'w')

def check_composite(n, a, d, s):
    x = pow(a, d, n)
    if x==1 or x==n-1:
        return False
    for r in range(1, s):
        x = x * x % n
        if x==n-1:
            return False
    return True

def miller_rabin(n):
    if n<2:
        return False
    r = 0 
    d = n-1
    while((d&1)==0):
        d >>=1
        r += 1
    base = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
    for a in base:
        if n==a:
            return True
        if check_composite(n, a, d, r):
            return False
    return True


n = inp()
a = 4 
b = a+n 
while(True):
    if not miller_rabin(a) and not miller_rabin(b):
        print(max(a,b),min(a,b))
        break
    else:
        a+=1
        b+=1