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 <bitset>
#include <cmath>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stdlib.h>
#include <string>
#include <utility>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define loop(i, x, n) for (int i = (x); i < (n); i++)
#define all(v) (v).begin(), (v).end()
#define int long long
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 1e9;
template<typename T> void cmax(T &a, T b) { a = max(a, b); }
template<typename T> void cmin(T &a, T b) { a = min(a, b); }
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

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

  if (b - a <= 2) {
    cout << 1 + k << endl;
  } else {
    int K = k - a + 1;
    cout << max(k + 1, a + K % 2 + K / 2 * (b - a)) << endl;
  }

  return 0;
}