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
#!/usr/bin/python3
import math

A = [0]*250

A[0] , A[1] = 0 , 45

for i in range(2,250):
	A[i] = A[i-1]*10 + 45*pow(10,i-1)

def tot(n):
	if n < 10: return (n*(n+1))//2
	d = len(str(n))-1
	p = pow(10,d)
	msd = n//p
	return msd*A[d] + ((msd*(msd-1))//2)*p + msd*(1+n%p) + tot(n%p)



a = int(input())
lo , hi = 1 , a


while lo < hi:
	mid = (lo+hi)>>1
	if tot(mid) >= a: hi = mid
	else : lo = mid+1

i1 , i2 = 1 , hi
total = tot(hi)

def digSum(n):
	total = 0
	while n>0:
		total += n%10
		n //= 10
	return total

while total%a != 0:
	if total < a:
		i2 += 1
		total += digSum(i2)
	else:
		total -= digSum(i1)
		i1 += 1
print(i1,i2)
	
	