#include <bits/stdc++.h>
using namespace std;
long long a, l = 0, r = 0;
long long sl = 0, sr = 0;
inline int calc(long long x) { return x ? calc(x / 10) + x % 10 : 0; }
void dfs(int s, long long d, int dd) {
  if (dd < 0) return;
  int i;
  for (i = 0; i <= 9; i++)
    if (sr + (s + i) * d + 45 * dd * (d / 10) <= a)
      sr += (s + i) * d + 45 * dd * (d / 10);
    else
      break;
  r = r * 10 + i - (!dd);
  dfs(s + i, d / 10, dd - 1);
}
int main() {
  scanf("%I64d", &a);
  dfs(0, 1e16, 16);
  while (sr - sl != a) {
    while (sr - sl < a) sr += calc(++r);
    while (sr - sl > a) sl += calc(++l);
  }
  printf("%I64d %I64d\n", l + 1, r);
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long a, l = 0, r = 0;
long long sl = 0, sr = 0;
inline int calc(long long x) { return x ? calc(x / 10) + x % 10 : 0; }
void dfs(int s, long long d, int dd) {
  if (dd < 0) return;
  int i;
  for (i = 0; i <= 9; i++)
    if (sr + (s + i) * d + 45 * dd * (d / 10) <= a)
      sr += (s + i) * d + 45 * dd * (d / 10);
    else
      break;
  r = r * 10 + i - (!dd);
  dfs(s + i, d / 10, dd - 1);
}
int main() {
  scanf("%I64d", &a);
  dfs(0, 1e16, 16);
  while (sr - sl != a) {
    while (sr - sl < a) sr += calc(++r);
    while (sr - sl > a) sl += calc(++l);
  }
  printf("%I64d %I64d\n", l + 1, r);
  return 0;
}

```