#include <bits/stdc++.h>
using namespace std;
long long dp[20][200], s[20], b, a;
int lee(long long x) {
  int r = 0;
  while (x) {
    s[++r] = x % 10;
    x = x / 10;
  }
  return r;
}
long long dfs(long long len, int now, int lim) {
  if (!lim && dp[len][now] != -1) return dp[len][now];
  if (len == 0) return now;
  int u = 9;
  if (lim) u = s[len];
  long long sum = 0;
  for (int i = 0; i <= u; i++) {
    sum = (sum + dfs(len - 1, now + i, lim && i == u)) % a;
  }
  if (!lim) dp[len][now] = sum;
  return sum;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%lld", &a);
  int u = lee(1000000000000000000);
  b = dfs(u, 0, 1) % a;
  printf("%lld %lld\n", a - b + 1, a - b + 1000000000000000000);
  return 0;
}

### Prompt

Your task is to create 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;
long long dp[20][200], s[20], b, a;
int lee(long long x) {
  int r = 0;
  while (x) {
    s[++r] = x % 10;
    x = x / 10;
  }
  return r;
}
long long dfs(long long len, int now, int lim) {
  if (!lim && dp[len][now] != -1) return dp[len][now];
  if (len == 0) return now;
  int u = 9;
  if (lim) u = s[len];
  long long sum = 0;
  for (int i = 0; i <= u; i++) {
    sum = (sum + dfs(len - 1, now + i, lim && i == u)) % a;
  }
  if (!lim) dp[len][now] = sum;
  return sum;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%lld", &a);
  int u = lee(1000000000000000000);
  b = dfs(u, 0, 1) % a;
  printf("%lld %lld\n", a - b + 1, a - b + 1000000000000000000);
  return 0;
}

```