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<iostream>
#include<cstdio>
#include<cstring>
#include <cstdlib>  
#include <math.h>
#include <cmath>
#include<cctype>
#include<string>
#include<set>
#include<iomanip>
#include <map>
#include<algorithm>
#include <functional>
#include<vector>
#include<climits>
#include<stack>
#include<queue>
#include<bitset>
#include <deque>
#include <climits>
#include <typeinfo>
#include <utility> 
using namespace std;
using ll = long long;
template<typename T>using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const ll inf = 1LL << 60;
#define all(x) (x).begin(),(x).end()
#define puts(x) cout << x << endl
#define rep(i,m,n) for(ll i = m;i < n;++i)
#define pb push_back
#define fore(i,a) for(auto &i:a)
#define rrep(i,m,n) for(ll i = m;i >= n;--i)
#define INF INT_MAX/2

int main() {
	ll k, a, b;
	cin >> k >> a >> b;

	ll s = k - a + 1;
	if (k <= a - 1 || b - a <= 2)puts(k + 1);
	else puts(a + (s/2) * (b - a) + (s%2));
	return 0;
}