#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
mt19937_64 rngll(rng());
unsigned long long lim = 2e16, sm = 1.1e18;
unsigned long long p[19];
unsigned long long solve(unsigned long long x) {
  x++;
  unsigned long long ans = 0;
  for (int i = 16; i >= 0; i--) {
    unsigned long long d = x / p[i] % 10;
    ans += d * (d - 1) / 2 * p[i] + d * i * 9 * p[i] / 2 + d * (x % p[i]);
  }
  return ans;
}
int main() {
  unsigned long long m;
  cin >> m;
  p[0] = 1;
  for (unsigned long long i = 1; i < 19; i++) p[i] = p[i - 1] * 10;
  while (1) {
    unsigned long long x = rngll() % lim;
    unsigned long long v = solve(x);
    unsigned long long nxt = sm / m * m + v;
    unsigned long long cur = 0;
    for (unsigned long long i = 1LL << 54; i; i >>= 1) {
      if (solve(cur + i) <= nxt) cur += i;
    }
    if (solve(cur) == nxt) {
      printf("%lld %lld", x + 1, cur);
      return 0;
    }
  }
}

### Prompt

Your challenge is to write a Cpp solution 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;
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
mt19937_64 rngll(rng());
unsigned long long lim = 2e16, sm = 1.1e18;
unsigned long long p[19];
unsigned long long solve(unsigned long long x) {
  x++;
  unsigned long long ans = 0;
  for (int i = 16; i >= 0; i--) {
    unsigned long long d = x / p[i] % 10;
    ans += d * (d - 1) / 2 * p[i] + d * i * 9 * p[i] / 2 + d * (x % p[i]);
  }
  return ans;
}
int main() {
  unsigned long long m;
  cin >> m;
  p[0] = 1;
  for (unsigned long long i = 1; i < 19; i++) p[i] = p[i - 1] * 10;
  while (1) {
    unsigned long long x = rngll() % lim;
    unsigned long long v = solve(x);
    unsigned long long nxt = sm / m * m + v;
    unsigned long long cur = 0;
    for (unsigned long long i = 1LL << 54; i; i >>= 1) {
      if (solve(cur + i) <= nxt) cur += i;
    }
    if (solve(cur) == nxt) {
      printf("%lld %lld", x + 1, cur);
      return 0;
    }
  }
}

```