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 JAVA):

Solution:
//package com.company;
import java.util.*;
public class Main {

    public static void main(String[] args) {
        // write your code here

        Scanner s=new Scanner(System.in);StringBuilder sb=new StringBuilder();
long n=s.nextLong();
if(n%2==0){
    System.out.println((4+n)+" "+4);
}else{
    System.out.println((9+n)+" "+9);
}
}
}


 class Student
{
    int rollno,x,pos;
//    String name, address;

    // Constructor
    public Student(int rollno, int x,int pos)
    {
        this.rollno = rollno;
      this.x=x;this.pos=pos;
    }

    // Used to print student details in main()
    public String toString()
    {
        return this.rollno + " " +this.x;
    }
}

 class Sortbyroll implements Comparator<Student>
{
    // Used for sorting in ascending order of
    // roll number
    public int compare(Student a, Student b)
    {
        return  b.x-a.x;
    }
}