#include <bits/stdc++.h>
using namespace std;
unsigned long long a, l, r, p;
unsigned long long read() {
  unsigned long long X = 0, F = 1;
  char C = getchar();
  while (C < '0' || C > '9') {
    if (C == '-') F = -1;
    C = getchar();
  }
  while (C >= '0' && C <= '9') {
    X = X * 10 + C - '0';
    C = getchar();
  }
  return X * F;
}
unsigned long long ksm(unsigned long long base, unsigned long long k,
                       unsigned long long mod) {
  base %= mod;
  unsigned long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * base % mod;
    base = base * base % mod;
    k >>= 1;
  }
  return ans;
}
int main() {
  a = read();
  p = ksm(10, 18, a) % a * 9 % a * 9 % a;
  l = a - p;
  r = 1000000000000000000 - p - 1 + a;
  printf("%llu %llu\n", l, r);
  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;
unsigned long long a, l, r, p;
unsigned long long read() {
  unsigned long long X = 0, F = 1;
  char C = getchar();
  while (C < '0' || C > '9') {
    if (C == '-') F = -1;
    C = getchar();
  }
  while (C >= '0' && C <= '9') {
    X = X * 10 + C - '0';
    C = getchar();
  }
  return X * F;
}
unsigned long long ksm(unsigned long long base, unsigned long long k,
                       unsigned long long mod) {
  base %= mod;
  unsigned long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * base % mod;
    base = base * base % mod;
    k >>= 1;
  }
  return ans;
}
int main() {
  a = read();
  p = ksm(10, 18, a) % a * 9 % a * 9 % a;
  l = a - p;
  r = 1000000000000000000 - p - 1 + a;
  printf("%llu %llu\n", l, r);
  return 0;
}

```