CF1866A.Ambitious Kid

Chaneka, Pak Chanek's child, is an ambitious kid, so Pak Chanek gives her the following problem to test her ambition.

Given an array of integers $[A_1, A_2, A_3, \ldots, A_N]$. In one operation, Chaneka can choose one element, then increase or decrease the element's value by $1$. Chaneka can do that operation multiple times, even for different elements.

What is the minimum number of operations that must be done to make it such that $A_1 \times A_2 \times A_3 \times \ldots \times A_N = 0$?

Chaneka 是 Pak Chanek 的孩子，是一位雄心勃勃的小孩，因此 Pak Chanek 给了她如下问题来测试她的雄心。

给定一个整数数组 $[A_1, A_2, A_3, \ldots, A_N]$。在一次操作中，Chaneka 可以选择其中一个元素，并将其值增加或减少 $1$。Chaneka 可以执行该操作多次，甚至对不同的元素执行。

为使 $A_1 \times A_2 \times A_3 \times \ldots \times A_N = 0$ 成立，所需的最少操作次数是多少？

The first line contains a single integer $N$ ($1 \leq N \leq 10^5$).

The second line contains $N$ integers $A_1, A_2, A_3, \ldots, A_N$ ($-10^5 \leq A_i \leq 10^5$).

第一行包含一个整数 $N$（$1 \leq N \leq 10^5$）。

第二行包含 $N$ 个整数 $A_1, A_2, A_3, \ldots, A_N$（$-10^5 \leq A_i \leq 10^5$）。

An integer representing the minimum number of operations that must be done to make it such that $A_1 \times A_2 \times A_3 \times \ldots \times A_N = 0$.

一个整数，表示使 $A_1 \times A_2 \times A_3 \times \ldots \times A_N = 0$ 成立所需的最少操作次数。

In the first example, initially, $A_1\times A_2\times A_3=2\times(-6)\times5=-60$. Chaneka can do the following sequence of operations:

1. Decrease the value of $A_1$ by $1$. Then, $A_1\times A_2\times A_3=1\times(-6)\times5=-30$
2. Decrease the value of $A_1$ by $1$. Then, $A_1\times A_2\times A_3=0\times(-6)\times5=0$

In the third example, Chaneka does not have to do any operations, because from the start, it already holds that $A_1\times A_2\times A_3\times A_4\times A_5=0\times(-1)\times0\times1\times0=0$

在第一个例子中，初始时，$A_1\times A_2\times A_3=2\times(-6)\times5=-60$。Chaneka 可以执行以下操作序列：

1. 将 $A_1$ 的值减小 $1$。此时，$A_1\times A_2\times A_3=1\times(-6)\times5=-30$
2. 将 $A_1$ 的值减小 $1$。此时，$A_1\times A_2\times A_3=0\times(-6)\times5=0$

在第三个例子中，Chaneka 无需执行任何操作，因为初始时已满足 $A_1\times A_2\times A_3\times A_4\times A_5=0\times(-1)\times0\times1\times0=0$