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
#include <bits/stdc++.h>
using namespace std;
char BUFFER[100000 + 5];
bool readn(int &n) { return scanf("%d", &n) == 1; }
bool readl(long long &n) { return scanf("%I64d", &n) == 1; }
bool readd(double &n) { return scanf("%lf", &n) == 1; }
bool reads(string &s) {
  s = "";
  int n = scanf("%s", BUFFER);
  if (n == 1) s = BUFFER;
  return n == 1;
}
bool readln(string &s) {
  char *valid = gets(BUFFER);
  if (valid) s = BUFFER;
  return ((bool)valid);
}
int Print(long long a, long long b) {
  cout << a << " " << b;
  return 0;
}
void update(long long &su, long long n, long long mod) {
  su *= n;
  su %= mod;
}
int doit(long long n, long long mod) {
  long long su = 1;
  int i;
  for (i = 1; i < n; i++) {
    su *= 10;
    su %= mod;
  }
  update(su, 3, mod);
  update(su, 3, mod);
  update(su, 5, mod);
  update(su, 2, mod);
  update(su, 3, mod);
  update(su, 3, mod);
  su++;
  su %= mod;
  long long a = 1, b = 1;
  for (i = 0; i < n; i++) b *= 10;
  if (0 == su) return Print(a, b);
  su = mod - su;
  return Print(a + su, b + su);
}
int main() {
  long long a;
  cin >> a;
  doit(18, a);
  printf("\n");
  return 0;
}
