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 maxn = 1.2e7 + 10;
int pri[maxn << 1];
int tot;
bool vis[maxn];
void pre() {
  for (int i = 2; i < maxn; i++) {
    if (!vis[i]) {
      pri[tot++] = i;
    }
    for (int j = 0; j < tot; j++) {
      if (i * pri[j] >= maxn) break;
      vis[i * pri[j]] = true;
      if (i % pri[j] == 0) break;
    }
  }
}
bool check(int n) {
  for (int i = 0; i < tot && pri[i] * pri[i] <= n; i++) {
    if (n % pri[i] == 0) return true;
  }
  return false;
}
int main() {
  pre();
  int n;
  scanf("%d", &n);
  for (int i = 1; i < maxn; i++) {
    if (vis[i] && n + i < maxn && vis[n + i]) {
      printf("%d %d\n", n + i, i);
      return 0;
    } else if (vis[i] && check(n + i)) {
      printf("%d %d\n", n + i, i);
      return 0;
    }
  }
}