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;
bool a[10000001];
int main() {
  long long int n = 10000000;
  for (int i = 0; i <= n; ++i) {
    a[i] = false;
  }
  for (long long int i = 3; i <= n; i = i + 2) {
    if (a[i] == false) {
      for (long long int j = i * i; j <= n; j = j + (2 * i)) {
        a[j] = true;
      }
    }
  }
  a[0] = true;
  a[1] = true;
  a[2] = false;
  long long int o;
  cin >> o;
  if (o % 2 == 0) {
    if (o == 2) {
      cout << 6 << " " << 4 << endl;
      return 0;
    }
    cout << o * 2 << " " << o << endl;
    return 0;
  } else {
    if (o == 1) {
      cout << 9 << " " << 8 << endl;
      return 0;
    }
    if (a[o] == true) {
      cout << o * 2 << " " << o << endl;
      return 0;
    } else {
      for (int i = o; i <= n; i = i + 2) {
        if (a[i] == true) {
          cout << (o * 2) + (i - o) << " " << i << endl;
          return 0;
        }
      }
    }
  }
  return 0;
}