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): 
  
    # Corner cases 
    if (n <= 1): 
        return False
    if (n <= 3): 
        return False
  
    # This is checked so that we can skip  
    # middle five numbers in below loop 
    if (n % 2 == 0 or n % 3 == 0): 
        return True
    i = 5
    while(i * i <= n): 
          
        if (n % i == 0 or n % (i + 2) == 0): 
            return True
        i = i + 6
          
    return False

n=int(input())
for b in range(2,10**9):
    if isComposite(b):
        a=b+n
        if isComposite(a):
            print(a,b)
            break