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 defaultdict as dc
'''from collections import deque as dq
from bisect import bisect_left,bisect_right,insort_left'''
import sys
import math
#define of c++ as inl=input()
mod=10**9 +7
def bs(a,x):
    i=bisect_left(a,x)
    if i!=len(a):
        return i
    else:
        return len(a)
def bs(a,b):
    l=a
    r=b+1
    x=b
    ans=0
    while(l<r):
        mid=(l+r)//2
        if x|mid>ans:
            ans=x|mid
            l=mid+1
        else:
            r=mid
    return ans
def digit(n):
    a=[]
    while(n>0):
        a.append(n%10)
        n=n//10
    return a
def inp():
    p=int(input())
    return p
def line():
    p=list(map(int,input().split()))
    return p
        
a=inp()
if a==1:
    print(9,8)
else:
    print(4*a,3*a)