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 Ruby (2.3.3) (judge verdict: Accepted)

k,a,b = gets.split.map &:to_i

#操作はk回
#1枚ビスケットを増やす
#a枚のビスケットと1円を交換
#1円とb枚のビスケットを交換
#最初は1枚ビスケットを持っている

#お金を経由するかどうか

if a + 2 >= b
    puts k + 1
    exit
end

#a-1回叩く
#k-(a-1)回の操作でAとBの操作を行う

#k-(a-1)が偶数
#(k-(a-1))/2回(b-a)個のビスケットが増える
if (k-(a-1)) % 2 == 0
    puts a + (b-a) * (k-(a-1)) / 2
    exit
else
    #k-(a-1)が奇数のときはもう1回叩く
    puts a + 1 + (b-a) * (k-(a)) / 2
    exit
end