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;
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
long long int binarysearch(long long int a[], long long int val,
                           long long int n) {
  long long int l = 0;
  long long int r = n - 1;
  long long int pos = abs(val);
  long long int mid;
  while (l < r) {
    mid = (l + r) / 2;
    if (a[mid] > pos)
      r = mid;
    else
      l = mid;
    if (r - l == 1) break;
  }
  if (a[l] > pos)
    return n - l;
  else if (a[r] > pos)
    return n - r;
  else
    return 0;
}
void answer(long long int n, long long int a[]) {
  if (a[n / 2] == 9) {
    a[n / 2] = 0;
    for (long long int i = n / 2 - 1; i >= 0; i--) {
      if (i == 0 && a[i] == 9) {
        a[i]++;
        break;
      }
      if (a[i] == 9) {
        a[i] = 0;
        continue;
      } else {
        a[i]++;
        break;
      }
    }
  }
  for (long long int i = 0; i <= n / 2; i++) {
    cout << a[i] << " ";
  }
  for (long long int i = n / 2 - 1; i >= 0; i--) cout << a[i] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  if (n % 2 == 0)
    cout << 4 + n << " " << 4;
  else
    cout << 9 + n << " " << 9;
  return 0;
}