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>
#pragma GCC optimize "trapv"
using namespace std;
const int N = 1e+5;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res % p;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t, n, b, a, s[N], ans;
  t = 1;
  while (t--) {
    cin >> n;
    if (n & 1)
      cout << 9 + n << ' ' << "9";
    else
      cout << 8 + n << ' ' << "8";
  }
}