Problem:
Score : 200 points

Problem Statement

You have N apples, called Apple 1, Apple 2, Apple 3, ..., Apple N. The flavor of Apple i is L+i-1, which can be negative.

You can make an apple pie using one or more of the apples. The flavor of the apple pie will be the sum of the flavors of the apples used.

You planned to make an apple pie using all of the apples, but being hungry tempts you to eat one of them, which can no longer be used to make the apple pie.

You want to make an apple pie that is as similar as possible to the one that you planned to make. Thus, you will choose the apple to eat so that the flavor of the apple pie made of the remaining N-1 apples will have the smallest possible absolute difference from the flavor of the apple pie made of all the N apples.

Find the flavor of the apple pie made of the remaining N-1 apples when you choose the apple to eat as above.

We can prove that this value is uniquely determined.

Constraints

2 \leq N \leq 200

-100 \leq L \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N L

Output

Find the flavor of the apple pie made of the remaining N-1 apples when you optimally choose the apple to eat.

Sample Input 1

5 2

Sample Output 1

18

The flavors of Apple 1, 2, 3, 4, and 5 are 2, 3, 4, 5, and 6, respectively. The optimal choice is to eat Apple 1, so the answer is 3+4+5+6=18.

Sample Input 2

3 -1

Sample Output 2

0

The flavors of Apple 1, 2, and 3 are -1, 0, and 1, respectively. The optimal choice is to eat Apple 2, so the answer is (-1)+1=0.

Sample Input 3

30 -50

Sample Output 3

-1044

Solution in Python (3.4.3) (judge verdict: Accepted)

# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import random
input = sys.stdin.readline

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

def main():

	#S = int(input())
	#S = list(input())
	N,L = map(int, input().split(" "))
	#L = list(map(int, input().split(" ")))

	azi = []
	for i in range(1, N+1):
		azi.append(i+L-1)

	tmp = bisect_left(azi, 0)

	ans = 0
	flg = False
	for i,val in enumerate(azi):
		if i == tmp:
			flg = True
			continue
		ans += val
	
	if flg == False:
		ans -= azi[-1]
	print(ans)

if __name__ == "__main__":
	main()