Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.

Vasily has a candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make b went out candles into a new candle. As a result, this new candle can be used like any other new candle.

Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.

Input

The single line contains two integers, a and b (1 ≤ a ≤ 1000; 2 ≤ b ≤ 1000).

Output

Print a single integer — the number of hours Vasily can light up the room for.

Examples

Input

4 2


Output

7


Input

6 3


Output

8

Note

Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
# Lets goto the next level 
# AIM Specialist at CF *__* asap 
# template taken from chaudhary_19  
# Remember you were also a novice when you started, 
# hence never be rude to anyone who wants to learn something
# Never open a ranklist untill and unless you are done with solving problems, wastes 3/4 minuts 
# Donot treat CP as a placement thing, love it and enjoy it, you will succeed for sure.  
# Any doubts or want to have a talk, contact https://www.facebook.com/chaudhary.mayank


# ///==========Libraries, Constants and Functions=============///

import sys
from bisect import bisect_left,bisect_right,insort
from collections import deque,Counter
from math import gcd,sqrt,factorial,ceil,log10,log2
from itertools import permutations
from heapq import heappush,heappop,heapify
inf = float("inf")
mod = 1000000007
#sys.setrecursionlimit(10**5)


def factorial_p(n, p):
    ans = 1
    if n <= p // 2:
        for i in range(1, n + 1):
            ans = (ans * i) % p
    else:
        for i in range(1, p - n):
            ans = (ans * i) % p
        ans = pow(ans, p - 2, p)
        if n % 2 == 0:
            ans = p - ans
    return ans


def nCr_p(n, r, p):
    ans = 1
    while (n != 0) or (r != 0):
        a, b = n % p, r % p
        if a < b:
            return 0
        ans = (ans * factorial_p(a, p) * pow(factorial_p(b, p), p - 2, p) * pow(factorial_p(a - b, p), p - 2, p)) % p
        n //= p
        r //= p
    return ans

def prime_sieve(n):
    """returns a sieve of primes >= 5 and < n"""
    flag = n % 6 == 2
    sieve = bytearray((n // 3 + flag >> 3) + 1)
    for i in range(1, int(n**0.5) // 3 + 1):
        if not (sieve[i >> 3] >> (i & 7)) & 1:
            k = (3 * i + 1) | 1
            for j in range(k * k // 3, n // 3 + flag, 2 * k):
                sieve[j >> 3] |= 1 << (j & 7)
            for j in range(k * (k - 2 * (i & 1) + 4) // 3, n // 3 + flag, 2 * k):
                sieve[j >> 3] |= 1 << (j & 7)
    return sieve
def prime_list(n):
    """returns a list of primes <= n"""
    res = []
    if n > 1:
        res.append(2)
    if n > 2:
        res.append(3)
    if n > 4:
        sieve = prime_sieve(n + 1)
        res.extend(3 * i + 1 | 1 for i in range(1, (n + 1) // 3 + (n % 6 == 1)) if not (sieve[i >> 3] >> (i & 7)) & 1)
    return res
    
        
def binary(number):                                             #  <----- calculate the no. of 1's in binary representation of number
    result=0
    while number:
        result=result+1
        number=number&(number-1)
    return result

def is_prime(n):
    """returns True if n is prime else False"""
    if n < 5 or n & 1 == 0 or n % 3 == 0:
        return 2 <= n <= 3
    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            p = (p * p) % n
            if p == n - 1:
                break
        else:
            return False
    return True

def all_factors(n):
    """returns a sorted list of all distinct factors of n"""
    small, large = [], []
    for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
        if not n % i:
            small.append(i)
            large.append(n // i)
    if small[-1] == large[-1]:
        large.pop()
    large.reverse()
    small.extend(large)
    return small

def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
# ///===========MAIN=============///
a,b=get_ints()
if b>a:
    print(a)
elif a==b:
    print(a+1)
else:
    total=a
    while(a//b>=1):
        extra=a%b
        a-=extra
        total+=a//b
        a=a//b+extra
    print(total)
