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;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string& s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cerr << "\033[33;0m" << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
template <typename T>
void debug_out(T a[], int n) {
  cerr << "{" << to_string(a[0]);
  for (int i = 1; i < n; ++i) cerr << ", " << to_string(a[i]);
  cerr << "}";
  debug_out();
}
using LL = long long;
using PII = pair<int, int>;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
LL powmod(LL a, LL b, LL MOD) {
  LL ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % MOD;
    b >>= 1;
    a = a * a % MOD;
  }
  return ans % MOD;
}
const int N = 100005, MOD = 1e9 + 7, INF = 1e9;
bool chk(int x) {
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return 1;
  }
  return 0;
}
void init() {
  int n;
  cin >> n;
  while (1) {
    int x = rand() % 10000 + 1;
    if (chk(x) && chk(x + n)) {
      cout << x + n << " " << x << endl;
      exit(0);
    }
  }
}
void work() {}
int main() {
  time_t tic = clock();
  int T = 1;
  while (T--) {
    init();
    work();
  }
  time_t toc = clock();
  cerr << "execute time = " << (toc - tic) / CLOCKS_PER_SEC << "ms\n";
  return 0;
}