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
#include <bits/stdc++.h>
using namespace std;
int f(long long p) {
  int ans = 0;
  while (p) {
    ans += p % 10;
    p /= 10;
  }
  return ans;
}
long long g(long long a) {
  if (a == 0) return 0;
  int digit = 0;
  long long b = a;
  while (b >= 10) {
    digit++;
    b /= 10;
  }
  int b_digit = b;
  long long ans = b * digit;
  for (int i = 0; i < digit; i++) ans *= 10, b *= 10;
  ans *= 9;
  ans /= 2;
  for (int i = 0; i < b_digit; i++) ans += i * (b / b_digit);
  ans += g(a - b) + b_digit * (a - b + 1);
  return ans;
}
int main() {
  long long a;
  scanf("%lld", &a);
  long long l = 1, r = 20000000000000000LL;
  while (l < r) {
    long long mid = (l + r) / 2;
    if (g(mid) >= a)
      r = mid;
    else
      l = mid + 1;
  }
  l = 1;
  long long cur = g(r);
  while (cur != a) {
    if (cur > a) {
      cur -= f(l);
      l++;
    } else {
      cur += f(r + 1);
      r++;
    }
  }
  printf("%lld %lld", l, r);
}
