#include <bits/stdc++.h>
using namespace std;
long long INF = 1e18;
template <class T>
inline void read(T &f) {
  f = 0;
  T x = 1;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') {
      x = -1;
    }
    ch = getchar();
  }
  while (ch <= 57 && ch >= 48) {
    f = f * 10 + ch - '0';
    ch = getchar();
  }
  f *= x;
}
template <class T>
void print(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    print(x / 10);
  }
  putchar(x % 10 + '0');
}
long long n;
long long ans;
int main() {
  read(n);
  ans = (INF % n * 3 % n * 3 % n * 3 % n * 3 % n + 1 % n) % n;
  print(n - ans + 1);
  cout << " ";
  print(INF + n - ans);
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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 INF = 1e18;
template <class T>
inline void read(T &f) {
  f = 0;
  T x = 1;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') {
      x = -1;
    }
    ch = getchar();
  }
  while (ch <= 57 && ch >= 48) {
    f = f * 10 + ch - '0';
    ch = getchar();
  }
  f *= x;
}
template <class T>
void print(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    print(x / 10);
  }
  putchar(x % 10 + '0');
}
long long n;
long long ans;
int main() {
  read(n);
  ans = (INF % n * 3 % n * 3 % n * 3 % n * 3 % n + 1 % n) % n;
  print(n - ans + 1);
  cout << " ";
  print(INF + n - ans);
  return 0;
}

```