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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<functional>
#include<iomanip>
#include<iostream>
#include<list>
#include<map>
#include<queue>
#include<random>
#include<set>
#include<stack>
#include<string>
#include<utility>
#include<vector>

#define FOR(i,a,b) for(long long i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define FORR(i,a,b) for(long long i=(a);i>=(b);--i)
#define REPR(i,n)   FORR(i,n,0)
#define int long long
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll>pll;

signed main() {
	int k,a,b;
    cin>>k>>a>>b;
	
    int money=0,bis=1;

    if(b<a+2||k+1<a+2){
        cout<<k+1<<endl;
        return 0;
    }

    int ans=0;
    k++;
    ans+=b;
    k-=(a+2);
    ans+=k/2*(b-a);
    ans+=k%2;

    cout<<ans<<endl;
	
	return 0;
}