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;
int SET(int n, int pos) { return n = n | (1 << pos); }
int RESET(int n, int pos) { return n = n & ~(1 << pos); }
int CHECK(int n, int pos) { return (bool)(n & (1 << pos)); }
int str2int(string s) {
  stringstream ss(s);
  int x;
  ss >> x;
  return x;
}
string int2str(int a) {
  stringstream ss;
  ss << a;
  string str = ss.str();
  return str;
}
string char2str(char a) {
  stringstream ss;
  ss << a;
  string str = ss.str();
  return str;
}
long long int bigMod(long long int n, long long int power, long long int MOD) {
  if (power == 0) return 1;
  if (power % 2 == 0) {
    long long int ret = bigMod(n, power / 2, MOD);
    return ((ret % MOD) * (ret % MOD)) % MOD;
  } else
    return ((n % MOD) * (bigMod(n, power - 1, MOD) % MOD)) % MOD;
}
long long int modInverse(long long int n, long long int MOD) {
  return bigMod(n, MOD - 2, MOD);
}
int POW(int x, int y) {
  int res = 1;
  for (; y;) {
    if ((y & 1)) {
      res *= x;
    }
    x *= x;
    y >>= 1;
  }
  return res;
}
int inverse(int x) {
  double p = ((double)1.0) / x;
  return (p) + 1e-9;
}
int gcd(int a, int b) {
  while (b) b ^= a ^= b ^= a %= b;
  return a;
}
int nC2(int n) { return n * (n - 1) / 2; }
long long int MOD(long long int n, long long int mod) {
  if (n >= 0)
    return n % mod;
  else if (-n == mod)
    return 0;
  else
    return mod + (n % mod);
}
int main() {
  int t, cas = 0;
  int n, b, p;
  cin >> n >> b >> p;
  int bot = 0, tow = n * p;
  while (n > 1) {
    int pp = 1;
    while (1) {
      if (pp * 2 > n) break;
      pp = pp * 2;
    }
    int xx = n - pp;
    pp /= 2;
    bot += ((2 * b) + 1) * pp;
    n = pp + xx;
  }
  cout << bot << " " << tow << endl;
  return 0;
}
