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:
def isComposite(n):
    i=5
    if (n <= 1) :
        return 0 
    if (n <= 3) :
        return 0 
    if ((n%2 == 0)or(n%3 == 0)) :
        return 1
    while (i*i<=n): 
        if ((n%i == 0)or(n%(i+2) == 0)): 
           return 1 
        i+=6   
    return 0
n=int(input())
list_pri=[]
list_pri.append(0)
for i in range(1,10**9):
    if(isComposite(i)and(isComposite(i+n))):
        print(i+n,i)
        break