Problem: zscoder wants to generate an input file for some programming competition problem.

His input is a string consisting of n letters 'a'. He is too lazy to write a generator so he will manually generate the input in a text editor.

Initially, the text editor is empty. It takes him x seconds to insert or delete a letter 'a' from the text file and y seconds to copy the contents of the entire text file, and duplicate it.

zscoder wants to find the minimum amount of time needed for him to create the input file of exactly n letters 'a'. Help him to determine the amount of time needed to generate the input.

Input

The only line contains three integers n, x and y (1 ≤ n ≤ 107, 1 ≤ x, y ≤ 109) — the number of letters 'a' in the input file and the parameters from the problem statement.

Output

Print the only integer t — the minimum amount of time needed to generate the input file.

Examples

Input

8 1 1


Output

4


Input

8 1 10


Output

8

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxpq = priority_queue<T>;
template <typename T>
void printv(const vector<T>& v) {
  for (auto i : v) cerr << i << ' ';
  cerr << '\n';
}
inline long long CC(long long n, long long y) {
  return (((n) + (y - 1) - 1) / (y - 1)) - (((n) + (y)-1) / (y));
}
inline long long FF(long long n, long long y) { return n / y - n / (y + 1); }
inline long long __gcd(long long a, long long b) {
  return !b ? a : __gcd(b, a % b);
}
inline int add(int a, int b, int p = 1000000007) {
  int c = a + b;
  if (c >= p) c -= p;
  return c;
}
inline int sub(int a, int b, int p = 1000000007) {
  int c = a - b;
  if (c < 0) c += p;
  return c;
}
inline int mul(int a, int b, int p = 1000000007) { return (a * 1ll * b) % p; }
long long n, dp[10000005], x, y;
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  clock_t clk = clock();
  cerr << "I will return...\n";
  cin >> n >> x >> y;
  dp[1] = x;
  for (long long i = (2); i < (n + 1); i++) {
    if (i % 2 == 0)
      dp[i] = min(dp[i - 1] + x, dp[i / 2] + y);
    else
      dp[i] = min(dp[i - 1] + x, dp[(i + 1) / 2] + x + y);
  }
  cout << dp[n] << '\n';
  cerr << "...don't you ever hang your head.\n";
  cerr << "Time (in ms): " << double(clock() - clk) * 1000.0 / CLOCKS_PER_SEC
       << '\n';
}