A tennis tournament with n participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out.

The tournament takes place in the following way (below, m is the number of the participants of the current round):

  * let k be the maximal power of the number 2 such that k ≤ m, 
  * k participants compete in the current round and a half of them passes to the next round, the other m - k participants pass to the next round directly, 
  * when only one participant remains, the tournament finishes. 



Each match requires b bottles of water for each participant and one bottle for the judge. Besides p towels are given to each participant for the whole tournament.

Find the number of bottles and towels needed for the tournament.

Note that it's a tennis tournament so in each match two participants compete (one of them will win and the other will lose).

Input

The only line contains three integers n, b, p (1 ≤ n, b, p ≤ 500) — the number of participants and the parameters described in the problem statement.

Output

Print two integers x and y — the number of bottles and towels need for the tournament.

Examples

Input

5 2 3


Output

20 15


Input

8 2 4


Output

35 32

Note

In the first example will be three rounds:

  1. in the first round will be two matches and for each match 5 bottles of water are needed (two for each of the participants and one for the judge), 
  2. in the second round will be only one match, so we need another 5 bottles of water, 
  3. in the third round will also be only one match, so we need another 5 bottles of water. 



So in total we need 20 bottles of water.

In the second example no participant will move on to some round directly.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int get_power(int n) {
  for (int i = 0; i < 20; i++) {
    if (pow(2, i) > n) {
      return pow(2, i - 1);
    }
  }
  return 0;
}
int main() {
  int n, b, p;
  cin >> n >> b >> p;
  int ans_b = 0, ans_p = n * p;
  while (n != 1) {
    int maxp = get_power(n);
    int match = maxp / 2;
    ans_b += (match * b * 2) + match;
    n = match + (n - match * 2);
    ;
  }
  cout << ans_b << " " << ans_p << "\n";
  return 0;
}
