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 Sieve(n): 
      
    
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
      
    l=list()    
    for p in range(2, n): 
        if prime[p]: 
            l.append(p)
    return l
   ''' 
a,b=0,0
#l=Sieve(1000000)
#l.append(1)
#print(l[:5])
n=int(input())
if n==1:
    a,b=9,8
elif n%2==0:
    a,b=n+4,4
else:
    a,b=3*n,2*n
print(a,b)