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;
const int N = 3e5 + 9, OO = 0x3f3f3f3f, MOD = 1000000007;
const long long OOLL = 0x3f3f3f3f3f3f3f3f;
int n, m, k;
bool prime(int x) {
  for (int i = 2; i <= x / i; ++i) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  int a = 1e9;
  int b = a - n;
  if (prime(b)) --a, --b;
  printf("%d %d\n", a, b);
  return 0;
}