Little X has met the following problem recently. 

Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>

Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code: 
    
    
      
        ans = solve(l, r) % a;  
        if (ans <= 0)  
          ans += a;  
      
    

This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.

Input

The first line contains a single integer a (1 ≤ a ≤ 1018).

Output

Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.

Examples

Input

46


Output

1 10


Input

126444381000032


Output

2333333 2333333333333
from random import *
pw=[1]
for i in range(1,40): pw.append(pw[i-1]*10)
def f(x):
	r=[]
	while x!=0:
		r.append(x%10)
		x/=10
	ans=0
	s=0
	r.reverse()
	for i in range(len(r)):
		if (i==len(r)-1): ans+=(2*s+r[i]-1)*r[i]/2
		else: ans+=(2*s+r[i]-1)*r[i]/2*pw[len(r)-i-1]+r[i]*45*(len(r)-i-1)*pw[len(r)-i-2]
		s+=r[i]
	return ans
mod=input()
pos=[-1]*1000
for i in range(2,40):
	pos[f(i)]=i
while True:
	x=randint(1,10**30)
	p=f(x)
	q=p%mod
	l,r=0,mod-q
	while l+1<r:
		md=(l+r)>>1
		if (f(x+md)-p<mod-q): l=md
		else: r=md
	P=f(x+r)
	Q=P%mod
	if pos[Q]!=-1 and pos[Q]!=x+r:
		a,b=pos[Q],x+r
		if a>b: a,b=b,a
		print a,b-1
		break
	pos[Q]=x+r