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;
const int N = 20;
const int LIMIT = 18;
int n;
long long g[N], a;
int f(int x) {
  if (x == 0) {
    return 0;
  } else {
    return x % 10 + f(x / 10);
  }
}
long long mul(long long c, int d) {
  long long res = 0;
  for (int i = 0; i < d; i++) {
    (res += c) %= a;
  }
  return res % a;
}
int main() {
  cin >> a;
  long long l = 1LL, r = 1LL;
  for (int i = 1; i <= LIMIT; i++) {
    r *= 10;
  }
  long long cur = 1, power = 1;
  for (int i = 1; i < LIMIT; i++) {
    power = mul(power, 10);
  }
  int digit = 45;
  power = mul(power, digit);
  power = mul(power, LIMIT);
  (++power) %= a;
  if (power == 0) {
    cout << l << " " << r << endl;
  } else {
    cout << l + a - power << " " << r + a - power << endl;
  }
  return 0;
}
