Problem:
Score : 400 points

Problem Statement

Snuke has one biscuit and zero Japanese yen (the currency) in his pocket.
He will perform the following operations exactly K times in total, in the order he likes:

Hit his pocket, which magically increases the number of biscuits by one.

Exchange A biscuits to 1 yen.

Exchange 1 yen to B biscuits.

Find the maximum possible number of biscuits in Snuke's pocket after K operations.

Constraints

1 \leq K,A,B \leq 10^9

K,A and B are integers.

Input

Input is given from Standard Input in the following format:

K A B

Output

Print the maximum possible number of biscuits in Snuke's pocket after K operations.

Sample Input 1

4 2 6

Sample Output 1

7

The number of biscuits in Snuke's pocket after K operations is maximized as follows:

Hit his pocket. Now he has 2 biscuits and 0 yen.

Exchange 2 biscuits to 1 yen. his pocket. Now he has 0 biscuits and 1 yen.

Hit his pocket. Now he has 1 biscuits and 1 yen.

Exchange 1 yen to 6 biscuits. his pocket. Now he has 7 biscuits and 0 yen.

Sample Input 2

7 3 4

Sample Output 2

8

Sample Input 3

314159265 35897932 384626433

Sample Output 3

48518828981938099

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder._20190209みんプロ
{
    class C
    {
        List<int> Convert(IEnumerable<string> str)
        {
            var res = new List<int>();
            foreach (var i in str)
            {
                var data = int.Parse(i);
                res.Add(data);
            }
            return res;
        }

        public void solve()
        {
            string input;
            while ((input = Console.ReadLine()) != null)
            {
                var data = Convert(input.Split(' '));
                var result = ANS(data[0], data[1], data[2]);
                Console.WriteLine(result);
            }
        }


        public long ANS(long k, long a, long b)
        {
            long res = 1;
            long diff = b - a;
            if (diff < 2)
            {
                res += k;
                return res;
            }

            for (int i = 0; i < k; i++)
            {
                if (res < a)
                {
                    res++;
                }
                else if (i == k - 1)
                {
                    res++;
                }
                else
                {
                    res += diff;
                    i++;
                }
            }

            return res;

        }


        static void Main(string[] args)
        {
            new C().solve();
        }
    }
}