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;
void init() {}
int binarySearch(long int arr[], int l, int r, long int x) {
  if (r >= l) {
    int mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return l;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int qwe;
  qwe = 1;
  init();
  while (qwe--) {
    long int n;
    cin >> n;
    if (n == 1)
      cout << 9 << " " << 8;
    else
      cout << n + n + n << " " << n + n;
  }
  return 0;
}