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:
#Bahadurgarh
n=(int)(input())

if n<1000000:
	prime = [True for i in range(1000000+1)] 
	p = 2
	while (p * p <= 1000000):
		if (prime[p] == True):
			for i in range(p * p, 1000000+1, p):
				prime[i] = False
		p += 1

	a=4
	b=4+n
	while prime[a]==True or prime[b]==True:
		a=a+1
		b=b+1
	
	print(b,a)
else:
	a=4
	b=4+n
	prime = [True for i in range(1000000+1)] 
	p = 2
	while (p * p <= 1000000):
		if (prime[p] == True):
			for i in range(p * p, 1000000+1, p):
				prime[i] = False
		p += 1	
	while True:
 		while prime[a]!=False:
 			a=a+1
 		b=a+n
 		i=2
 		while b%i!=0 and i<100000:
 			i=i+1
 		if i!=100000:
 			break
 		a=a+1
	print(b,a)