CF998B.Cutting

There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.

There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.

Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ two cuts $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$. On each segment the number of even elements should be equal to the number of odd elements.

The cost of the cut between $x$ and $y$ numbers is $|x - y|$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $B$ bitcoins.

有很多东西可以被“切”——树木、纸张、“绳子”。在本题中，你需要切割一个整数序列。

给定一个整数序列，其中奇数与偶数的个数相等。在预算有限的前提下，你需要进行尽可能多的切割操作，使得切割后得到的每个连续（即相邻）子段中奇数与偶数的个数均相等。

每次切割将序列划分为若干连续（contiguous）子段。你可以将每次切割理解为序列中两个相邻元素之间的断点。因此，切割完成后，每个元素恰好属于一个子段。例如：
$[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$ $\to$ 两次切割 $\to$ $[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$。
要求每个子段中偶数的个数等于奇数的个数。

在数字 $x$ 和 $y$ 之间进行一次切割的花费为 $|x - y|$ 比特币。求在总花费不超过 $B$ 比特币的前提下，最多能进行多少次切割。

First line of the input contains an integer $n$ ($2 \le n \le 100$) and an integer $B$ ($1 \le B \le 100$) — the number of elements in the sequence and the number of bitcoins you have.

Second line contains $n$ integers: $a_1$, $a_2$, ..., $a_n$ ($1 \le a_i \le 100$) — elements of the sequence, which contains the equal number of even and odd numbers

输入的第一行包含一个整数 $n$（$2 \le n \le 100$）和一个整数 $B$（$1 \le B \le 100$）——分别表示序列中元素的个数以及你拥有的比特币数量。

第二行包含 $n$ 个整数：$a_1$, $a_2$, ..., $a_n$（$1 \le a_i \le 100$）——序列的元素，该序列中偶数与奇数的个数相等。

Print the maximum possible number of cuts which can be made while spending no more than $B$ bitcoins.

输出在花费不超过 $B$ 比特币的前提下，最多可以进行的切割次数。

In the first sample the optimal answer is to split sequence between $2$ and $5$. Price of this cut is equal to $3$ bitcoins.

In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.

In the third sample the sequence should be cut between $2$ and $3$, and between $4$ and $5$. The total price of the cuts is $1 + 1 = 2$ bitcoins.

在第一个样例中，最优方案是在 $2$ 和 $5$ 之间将序列切开。该切割的代价为 $3$ 比特币。

在第二个样例中，即使拥有无限数量的比特币，也无法进行哪怕一次切割。

在第三个样例中，序列应在 $2$ 和 $3$ 之间以及 $4$ 和 $5$ 之间进行切割。切割的总代价为 $1 + 1 = 2$ 比特币。