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;
unsigned long long a, dp[20][20], mi[20], f[20], l, r;
int main() {
  cin >> a;
  mi[0] = 1;
  for (int i = 1; i <= 19; ++i) mi[i] = mi[i - 1] * 10;
  for (int i = 1; i <= 9; ++i) dp[0][i] = 1;
  for (int i = 1; i <= 19; ++i) {
    for (int j = 1; j <= 9; ++j)
      dp[i][j] = dp[i - 1][j] + dp[i - 1][j] * 9 + mi[i];
  }
  for (int i = 0; i <= 18; ++i) {
    for (int j = 1; j <= 9; ++j) {
      f[i + 1] = (f[i + 1] + dp[i][j] * j % a) % a;
    }
    f[i] = (f[i] + 1) % a;
  }
  for (int i = 0; i <= 19; ++i) {
    if (a - f[i] % a < mi[i]) {
      l = a - f[i] % a + 1;
      r = l + mi[i] - 1;
      break;
    }
  }
  cout << l << " " << r << endl;
  return 0;
}
