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;
using ll = long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<ld, ld>;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<ld>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
str Tout[2] = {"NO", "YES"};
str tout[2] = {"No", "Yes"};
int main() {
  int n, ans;
  cin >> n;
  if (n % 2)
    ans = 100000011;
  else
    ans = 100000010;
  cout << ans << " " << ans - n;
  return 0;
}