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 collections import Counter,deque,defaultdict,OrderedDict,namedtuple
from bisect import bisect_left,bisect_right,insort
from math import sqrt,ceil,floor,factorial,gcd

def p(n):
    d=0
    if n==1 or n==2:
        return True
    prev=1
    for i in range(2,int(sqrt(n)+1)):
        if n%i==0 and i!=prev:
            d+=1
            prev=i
        if d>0:
            break
    if d>0:
        return False
    else:
        return True

from sys import stdin    
n=int(stdin.readline())
a=4
b=0
while b==0:
    if p(a)==False:
        if p(n+a)==False:
            b=n+a
            break
    a+=1
print(b,a)