#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-12, pi = acosl(-1);
const long long inf = 1e16, mod = 1e9 + 7;
const int N = 2e6 + 10;
inline long long sum(long long n) {
  long long a[20], d = 0, p10 = 1;
  while (n) {
    a[d++] = n % 10;
    n /= 10;
    p10 *= 10;
  }
  p10 /= 10;
  long long ans = 0, s = 0;
  for (long long i = d; i--;) {
    ans += a[i] * p10 * (9 * i + 2 * s + a[i] - 1) / 2;
    s += a[i];
    p10 /= 10;
  }
  return ans;
}
inline long long digsum(long long n) {
  long long ans(0);
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int main() {
  long long a;
  cin >> a;
  long long l = 1, r = a, m, b;
  if (r > 1e15) r /= 10;
  while (l < r) {
    m = (l + r + 1) / 2;
    b = sum(m);
    if (b < a)
      l = m;
    else if (b > a)
      r = m - 1;
    else {
      cout << "1 " << m - 1;
      return 0;
    }
  }
  b = sum(l);
  r = l;
  l = 1;
  while (b != a) {
    if (b < a) b += digsum(r++);
    if (b > a) b -= digsum(l++);
  }
  cout << l << " " << r - 1;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const long double eps = 1e-12, pi = acosl(-1);
const long long inf = 1e16, mod = 1e9 + 7;
const int N = 2e6 + 10;
inline long long sum(long long n) {
  long long a[20], d = 0, p10 = 1;
  while (n) {
    a[d++] = n % 10;
    n /= 10;
    p10 *= 10;
  }
  p10 /= 10;
  long long ans = 0, s = 0;
  for (long long i = d; i--;) {
    ans += a[i] * p10 * (9 * i + 2 * s + a[i] - 1) / 2;
    s += a[i];
    p10 /= 10;
  }
  return ans;
}
inline long long digsum(long long n) {
  long long ans(0);
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int main() {
  long long a;
  cin >> a;
  long long l = 1, r = a, m, b;
  if (r > 1e15) r /= 10;
  while (l < r) {
    m = (l + r + 1) / 2;
    b = sum(m);
    if (b < a)
      l = m;
    else if (b > a)
      r = m - 1;
    else {
      cout << "1 " << m - 1;
      return 0;
    }
  }
  b = sum(l);
  r = l;
  l = 1;
  while (b != a) {
    if (b < a) b += digsum(r++);
    if (b > a) b -= digsum(l++);
  }
  cout << l << " " << r - 1;
  return 0;
}

```