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 = 1e6 + 5;
int pre[N], tot;
bool vis[N];
void init() {
  for (int i = 2; i <= N - 5; i++) {
    if (!vis[i]) pre[++tot] = i;
    for (int j = 1; j <= tot; j++) {
      if (i * pre[j] > N - 5) break;
      vis[i * pre[j]] = 1;
      if (i % pre[j] == 0) break;
    }
  }
}
bool judge(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return 1;
  }
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 4;; i += 2) {
    if (judge(i + n)) {
      printf("%d %d\n", i + n, i);
      return 0;
    }
  }
  return 0;
}