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 <bits/stdc++.h>

using namespace std;

using ll = long long;
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define FORE(i, a, b) for(int i=(a);i<=(b);++i)
#define FORR(i, a, b) for(int i=(b-1);i>=(a);--i)
#define FORRE(i, a, b) for(int i=(b);i>=(a);--i)
#define rep(i, n) for(int i=0;i<(n);++i)
#define repe(i, n) for(int i=0;i<=(n);++i)
#define repr(i, n) for(int i=(n-1);i>=0;--i)
#define repre(i, n) for(int i=(n);i>=0;--i)
#define ALL(v) (v).begin(),(v).end()
#define SP cout<<fixed<<setprecision(10)
typedef pair<int, int> P;
const int INF = (int) 1e9;
const int MOD = (int) 1e9 + 7;

int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

int main(void) {
	cin.tie(0);
	ios::sync_with_stdio(false);
	
	ll k, a, b;
	cin >> k >> a >> b;
	
	if ((a + 2) >= b or (k + 2) <= a) {
		cout << k + 1 << endl;
		return 0;
	}
	
	ll cnt = a;
	k -= a - 1;
	cnt += (b - a) * (k / 2);
	if (k % 2 == 1) cnt++;
	
	cout << cnt << endl;
	
	return 0;
}