#include <bits/stdc++.h>
using namespace std;
using std::bitset;
const double Pi = acos(-1);
inline int read() {
  int x = 0;
  bool fu = 0;
  char ch = 0;
  while (ch > '9' || ch < '0') {
    ch = getchar();
    if (ch == '-') fu = 1;
  }
  while (ch <= '9' && ch >= '0') x = (x * 10 - 48 + ch), ch = getchar();
  return fu ? -x : x;
}
inline char readc() {
  register char c;
  while ((c = getchar()) == ' ' || c == '\n')
    ;
  return c;
}
long long a;
long long x, y = 100000000000000000ll;
int main() {
  cin >> a;
  for (register int i = 1, iend = 18; i <= iend; ++i) {
    for (register int j = 1, jend = 9; j <= jend; ++j) {
      x += y * j;
      x %= a;
    }
  }
  x++;
  x = x % a;
  cout << (long long)(1 + a - x) << ' ' << (long long)(y * 10 + a - x);
  return 0;
}

### Prompt

Your challenge is to write 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;
using std::bitset;
const double Pi = acos(-1);
inline int read() {
  int x = 0;
  bool fu = 0;
  char ch = 0;
  while (ch > '9' || ch < '0') {
    ch = getchar();
    if (ch == '-') fu = 1;
  }
  while (ch <= '9' && ch >= '0') x = (x * 10 - 48 + ch), ch = getchar();
  return fu ? -x : x;
}
inline char readc() {
  register char c;
  while ((c = getchar()) == ' ' || c == '\n')
    ;
  return c;
}
long long a;
long long x, y = 100000000000000000ll;
int main() {
  cin >> a;
  for (register int i = 1, iend = 18; i <= iend; ++i) {
    for (register int j = 1, jend = 9; j <= jend; ++j) {
      x += y * j;
      x %= a;
    }
  }
  x++;
  x = x % a;
  cout << (long long)(1 + a - x) << ' ' << (long long)(y * 10 + a - x);
  return 0;
}

```