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 PyPy3 (2.4.0) (judge verdict: Accepted)

from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
import sys,bisect,string,math,time,functools,random
def Golf():*a,=map(int,open(0))
def I():return int(input())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def GI(V,E,Directed=False,index=0):
    org_inp=[];g=[[] for i in range(n)]
    for i in range(E):
        inp=LI();org_inp.append(inp)
        if index==0:inp[0]-=1;inp[1]-=1
        if len(inp)==2:
            a,b=inp;g[a].append(b)
            if not Directed:g[b].append(a)
        elif len(inp)==3:
            a,b,c=inp;aa=(inp[0],inp[2]);bb=(inp[1],inp[2]);g[a].append(bb)
            if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,boundary=1,search=[],replacement_of_found='.',mp_def={'#':1,'.':0}):
#h,w,g,sg=GGI(h,w,boundary=1,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0}) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in range(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def bit_combination(k,n=2):return [[tb//(n**bt)%n for bt in range(k)] for tb in range(n**k)]
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)
def show2d(g,h,w):
    for i in range(h):show(g[i*w:i*w+w])

YN=['YES','NO'];Yn=['Yes','No']
mo=10**9+7
inf=float('inf')
l_alp=string.ascii_lowercase
#sys.setrecursionlimit(10**7)
input=lambda: sys.stdin.readline().rstrip()

show_flg=False
show_flg=True




k,a,b=LI()

if a+1>=b:
    ans=k+1
else:
    if k<a:
        ans=1+k
    else:
        k-=a+1
        ans=b
        d=k//2
        ans+=(b-a)*d
        ans+=k%2

print(ans)