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 binario(f):
    valor=0
    if(f==1):
        return(1)
    for i in range(1,f+1):
        if(f%i==0):
            valor=valor+1
            if(valor>2):
                return(0)
                break
    if(valor==2):
        return(1)
n=int(input())
index=0;bucle=0
if(n%2==0):
    a=n+1
    b=a-n
else:
    a=n+2
    b=a-n
final=1
while(index==0):
    if(binario(a)==1):
        a=a+1
    elif(binario(b)==1):
        b=b+1
    if(binario(a)==0 and binario(b)==0):
        while(final==1):
            a=b+n
            if(binario(a)==0 and binario(b)==0):
                print(a,b)
                final=2
                index=5
            else:
                b=b*2