#include <bits/stdc++.h>
using namespace std;
const int bit = 100;
unsigned long long n, sum, len;
int main() {
  scanf("%llu", &n);
  sum = 45 * bit;
  for (int i = 1; i <= bit - 1; i++) sum = sum * 2 % n * 5 % n;
  sum = n - sum - 1;
  len = sum == 0 ? 1 : log10(sum) + 1;
  printf("%llu ", sum + 1);
  printf("1");
  for (int i = 1; i <= bit - len; i++) printf("0");
  printf("%llu\n", sum);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int bit = 100;
unsigned long long n, sum, len;
int main() {
  scanf("%llu", &n);
  sum = 45 * bit;
  for (int i = 1; i <= bit - 1; i++) sum = sum * 2 % n * 5 % n;
  sum = n - sum - 1;
  len = sum == 0 ? 1 : log10(sum) + 1;
  printf("%llu ", sum + 1);
  printf("1");
  for (int i = 1; i <= bit - len; i++) printf("0");
  printf("%llu\n", sum);
}

```