#include <bits/stdc++.h>
using namespace std;
long long need, tneed, ans;
long long l, r;
long long sadd(long long a, long long b) {
  if (a + b > 2e18) return 2e18;
  return a + b;
}
long long smult(long long a, long long b) {
  if (a == 0 || b == 0) return a * b;
  if (2e18 / b <= a) return 2e18;
  return a * b;
}
long digisum(long long x) {
  long res = 0;
  while (x) {
    res += x % 10;
    x /= 10;
  }
  return res;
}
long long worst;
long long solve(long long x) {
  if (x <= 0) return 0;
  long long res = 0;
  long long q, r;
  q = x / 10;
  r = x % 10;
  res = sadd(res, smult(q, 45ll));
  res = sadd(res, smult(10, solve(q - 1)));
  res = sadd(res, r * (r + 1) / 2);
  res = sadd(res, smult(r + 1, digisum(q)));
  return res;
}
long long gener() {
  long long res = 0;
  for (int i = 1; i <= 18; i++) res = res * 10 + rand() % 10;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  srand(13133);
  cin >> need;
  ans = -1;
  long long i = 0;
  while (true) {
    i++;
    l = i;
    r = 1e18;
    tneed = need + solve(l - 1);
    while (l < r) {
      long long mid = l + r;
      mid /= 2;
      if (solve(mid) >= tneed)
        r = mid;
      else
        l = mid + 1;
    }
    if (solve(l) == tneed) {
      ans = l;
      break;
    }
  }
  cout << i << " " << ans << endl;
  cin.get();
  cin.get();
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long need, tneed, ans;
long long l, r;
long long sadd(long long a, long long b) {
  if (a + b > 2e18) return 2e18;
  return a + b;
}
long long smult(long long a, long long b) {
  if (a == 0 || b == 0) return a * b;
  if (2e18 / b <= a) return 2e18;
  return a * b;
}
long digisum(long long x) {
  long res = 0;
  while (x) {
    res += x % 10;
    x /= 10;
  }
  return res;
}
long long worst;
long long solve(long long x) {
  if (x <= 0) return 0;
  long long res = 0;
  long long q, r;
  q = x / 10;
  r = x % 10;
  res = sadd(res, smult(q, 45ll));
  res = sadd(res, smult(10, solve(q - 1)));
  res = sadd(res, r * (r + 1) / 2);
  res = sadd(res, smult(r + 1, digisum(q)));
  return res;
}
long long gener() {
  long long res = 0;
  for (int i = 1; i <= 18; i++) res = res * 10 + rand() % 10;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  srand(13133);
  cin >> need;
  ans = -1;
  long long i = 0;
  while (true) {
    i++;
    l = i;
    r = 1e18;
    tneed = need + solve(l - 1);
    while (l < r) {
      long long mid = l + r;
      mid /= 2;
      if (solve(mid) >= tneed)
        r = mid;
      else
        l = mid + 1;
    }
    if (solve(l) == tneed) {
      ans = l;
      break;
    }
  }
  cout << i << " " << ans << endl;
  cin.get();
  cin.get();
  return 0;
}

```