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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 1e7 + 5;
bool notPrime[maxn];
void init() {
  memset(notPrime, 0, sizeof(notPrime));
  notPrime[0] = notPrime[1] = true;
  for (int i = 2; i < maxn; i++)
    for (int j = i * 2; j < maxn; j += i) notPrime[j] = true;
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  init();
  int n;
  cin >> n;
  for (int i = 4; i <= 10000000; i++) {
    if (notPrime[i + n] && notPrime[i]) {
      cout << i + n << " " << i << endl;
      break;
    }
  }
  return 0;
}