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>
using namespace std;
template <typename T>
T max(T &a, T &b) {
  return (!(a < b) ? a : b);
}
template <typename T>
T min(T &a, T &b) {
  return (a < b ? a : b);
}
template <typename T>
T mod(T a, T b) {
  return (a < b ? a : a % b);
}
long long mulmod(long long a, long long b, long long m) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T expo(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T power(T e, T n, T m) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mod(x * p, m);
    p = mod(p * p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T InverseEuler(T a, T m) {
  return (a == 1 ? 1 : power(a, m - 2, m));
}
template <typename T>
T gcd(T a, T b) {
  return (!b) ? a : __gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
template <typename T>
T bitpop(T x) {
  return __builtin_popcount(x);
}
long long n, b, p, n1, n2, tow, k = 0, bt;
long long facto2, facto1;
long long fact[1237LL + 1];
void pre() {}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> n;
  cin >> b;
  cin >> p;
  tow = p * n;
  bt = 0;
  n1 = n;
  n2 = n;
  while (n1 != 1) {
    k = 0;
    do {
      n2 = n2 >> 1;
      ++k;
    } while (n2 != 1);
    bt += (2 * b + 1) * (1 << (k - 1));
    n2 = n1 % (1 << k);
    n1 = expo(2LL, (k - 1)) + n2;
    n2 = n1;
  }
  cout << bt << ' ' << tow;
}
