#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T, class U>
inline void print(T first, U second) {
  printsp(first);
  println(second);
}
template <class T, class U, class V>
inline void print(T first, U second, V z) {
  printsp(first);
  printsp(second);
  println(z);
}
long long mul(long long a, long long b, long long c) {
  return ((long long)a * b - (long long)(a / (long double)c * b + 1e-4) * c +
          c) %
         c;
}
long long power(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a, c);
    a = mul(a, a, c);
    b >>= 1;
  }
  return ans;
}
int main() {
  long long a;
  gn(a);
  long long first = mul(45 * 18, power(10, 17, a), a);
  print(a - first, a - first + (long long)1e18 - 1);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T, class U>
inline void print(T first, U second) {
  printsp(first);
  println(second);
}
template <class T, class U, class V>
inline void print(T first, U second, V z) {
  printsp(first);
  printsp(second);
  println(z);
}
long long mul(long long a, long long b, long long c) {
  return ((long long)a * b - (long long)(a / (long double)c * b + 1e-4) * c +
          c) %
         c;
}
long long power(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a, c);
    a = mul(a, a, c);
    b >>= 1;
  }
  return ans;
}
int main() {
  long long a;
  gn(a);
  long long first = mul(45 * 18, power(10, 17, a), a);
  print(a - first, a - first + (long long)1e18 - 1);
  return 0;
}

```