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 JAVA):

Solution:
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.PriorityQueue;
public class E {
	public static void main(String args[]) {
		Sc sc = new Sc(System.in);
		long t = 0;
		int n = sc.nI();
		int x = sc.nI();
		int y = sc.nI();
		long[] time = new long[n+10];
		time[1] = x;
		time[2] = Math.min(2*x,x+y);
		for(int i = 4; i<=n+1; i+=2) {
			time[i] = time[i/2] + y;
			time[i-1] = Math.min(time[i] + x,time[i-2] + x);
			time[i] = Math.min(time[i],time[i-1] + x);
		}
		
		System.out.println(time[n]);
		
	}
}
class Sc {
    public Sc(InputStream i) {
        r = new BufferedReader(new InputStreamReader(i));
    }

    public boolean hasM() {
        return peekToken() != null;
    }

    public int nI() {
        return Integer.parseInt(nextToken());
    }

    public double nD() { 
        return Double.parseDouble(nextToken());
    }

    public long nL() {
        return Long.parseLong(nextToken());
    }

    public String n() {
        return nextToken();
    }

    private BufferedReader r;
    private String line;
    private StringTokenizer st;
    private String token;

    private String peekToken() {
        if (token == null) 
            try {
                while (st == null || !st.hasMoreTokens()) {
                    line = r.readLine();
                    if (line == null) return null;
                    st = new StringTokenizer(line);
                }
                token = st.nextToken();
            } catch (IOException e) { }
        return token;
    }

    private String nextToken() {
        String ans = peekToken();
        token = null;
        return ans;
    }
}