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 vl = vector<ll>;
using vi = vector<int>;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using ppii = pair<pi, int>;
using vvi = vector<vi>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpil = vector<pil>;
using vpli = vector<pli>;
using vppii = vector<ppii>;
using si = set<int>;
using sl = set<ll>;
using spi = set<pi>;
using spli = set<pli>;
int main() {
  int n;
  cin >> n;
  cout << 9 * n << " " << 8 * n;
}