#include <bits/stdc++.h>
using namespace std;
long long pw[20], dp[20];
long long g(long long x, long long a) {
  pw[0] = 1;
  dp[0] = 0;
  for (int i = 1; i <= 18; i++) {
    pw[i] = 10 * pw[i - 1];
    dp[i] =
        ((((((((45) % a) * ((i) % a)) % a)) % a) * ((pw[i - 1]) % a)) % a) % a;
  }
  long long sum = 0, ret = 0, flag = 0;
  for (int i = 18; i >= 1; i--) {
    long long d = (x / pw[i]) % 10;
    if (d > 0) flag = 1;
    if (!flag) continue;
    long long cur = 0;
    for (int j = 0; j < d; j++) {
      cur = ((((cur) % a) +
              ((((((((((((((sum) % a) + ((j) % a)) % a)) % a) * ((pw[i]) % a)) %
                     a)) %
                   a) +
                  ((dp[i]) % a)) %
                 a)) %
               a)) %
             a);
    }
    sum = ((((sum) % a) + ((d) % a)) % a);
    ret = ((((ret) % a) + ((cur) % a)) % a);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a;
  cin >> a;
  long long x = (45 * 18) % a;
  for (int i = 1; i < 18; i++) x = (x * 10) % a;
  long long l = a - x, r = ((long long)1e18) + a - x - 1;
  cout << l << " " << r << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 pw[20], dp[20];
long long g(long long x, long long a) {
  pw[0] = 1;
  dp[0] = 0;
  for (int i = 1; i <= 18; i++) {
    pw[i] = 10 * pw[i - 1];
    dp[i] =
        ((((((((45) % a) * ((i) % a)) % a)) % a) * ((pw[i - 1]) % a)) % a) % a;
  }
  long long sum = 0, ret = 0, flag = 0;
  for (int i = 18; i >= 1; i--) {
    long long d = (x / pw[i]) % 10;
    if (d > 0) flag = 1;
    if (!flag) continue;
    long long cur = 0;
    for (int j = 0; j < d; j++) {
      cur = ((((cur) % a) +
              ((((((((((((((sum) % a) + ((j) % a)) % a)) % a) * ((pw[i]) % a)) %
                     a)) %
                   a) +
                  ((dp[i]) % a)) %
                 a)) %
               a)) %
             a);
    }
    sum = ((((sum) % a) + ((d) % a)) % a);
    ret = ((((ret) % a) + ((cur) % a)) % a);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long a;
  cin >> a;
  long long x = (45 * 18) % a;
  for (int i = 1; i < 18; i++) x = (x * 10) % a;
  long long l = a - x, r = ((long long)1e18) + a - x - 1;
  cout << l << " " << r << endl;
  return 0;
}

```