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>
int main() {
  int n, b, p;
  std::cin >> n >> b >> p;
  std::vector<int> need_b(n, 0);
  std::queue<int> queue;
  for (int i = 0; i < n; ++i) {
    queue.push(i);
  }
  int calc_b = 0;
  while (queue.size() > 1u) {
    int pow2 = 1;
    while (pow2 * 2 <= int(queue.size())) {
      pow2 *= 2;
    }
    pow2 /= 2;
    std::queue<int> q2;
    for (int k = 0; k < pow2; ++k) {
      int i = queue.front();
      queue.pop();
      int j = queue.front();
      queue.pop();
      need_b[i] += b;
      need_b[j] += b;
      calc_b += 1;
      q2.push(i);
    }
    while (!queue.empty()) {
      q2.push(queue.front());
      queue.pop();
    }
    queue = q2;
  }
  calc_b += std::accumulate(need_b.begin(), need_b.end(), 0LL);
  std::cout << calc_b << " " << p * n << std::endl;
  return 0;
}
