Problem: Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:

  * the following numbers are composite: 1024, 4, 6, 9; 
  * the following numbers are not composite: 13, 1, 2, 3, 37. 



You are given a positive integer n. Find two composite integers a,b such that a-b=n.

It can be proven that solution always exists.

Input

The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.

Output

Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).

It can be proven, that solution always exists.

If there are several possible solutions, you can print any. 

Examples

Input


1


Output


9 8


Input


512


Output


4608 4096

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  int c = 0, v = 0;
  cin >> n;
  for (long long int i = 1; i < 1000000; i++) {
    long long int x = 0;
    v = 0;
    c = 0;
    for (long long int j = 2; j < i; j++) {
      if (i % j == 0) c++;
    }
    if (c > 0) {
      x = n + i;
      for (long long int u = 2; u < x; u++) {
        if (x % u == 0) v++;
      }
      if (v > 0) {
        cout << x << " " << i << "\n";
        break;
      }
    }
  }
  return 0;
}