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 <iomanip>
#include <algorithm>
#include <functional>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <string>
#include <list>
#include <math.h>
#include <string.h>
#include <bitset>
#include <queue>
using namespace std;

typedef pair<int, int> P;
#define INF (1LL<<60)
#define int long long
#define ll long long
#define mod 1000000007


signed main(){

	int K, A, B;
	cin >> K >> A >> B;
	int ans;
	if (A + 2 < B){
		if (K <= A) ans = 1 + K;
		else{
			ans = A + (K - (A - 1)) / 2 * (B - A) + (K - (A - 1)) % 2;
		}
	}
	else
		ans = 1 + K;

	cout << ans << endl;

	return 0;
}