西瓜Codeforces-4A

本文探讨了一个有趣的编程问题:如何判断能否将一个特定重量的西瓜均匀分成两份,每份重量均为偶数。此问题不仅考验对偶数的理解,还涉及编程逻辑与条件判断。

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Problem description

One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed w kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.

Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.

Input

The first (and the only) input line contains integer number w (1 ≤ w ≤ 100) — the weight of the watermelon bought by the boys.

Output

Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.

Examples

Input

8

Output

YES

Note

For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).

问题简述

将一个重w(1-100)千克的西瓜分成两部分,两部分的重量都要求是偶数。当西瓜的总重满足时,输出“YES”,反之,输出“NO”。

问题分析和程序说明

这道题只要读懂了题意,(even:偶数),其实不难,但如果要AC就要注意--仅仅是判断输入的w为偶数,还不行,因为忽略了偶数二的影响。题目没有要求多组数据输入,所以采用if语句来判断就好了。

另外,注意一下数学语言与计算机语言的差别:1<=w<=100是错误的,要改成1<=w&&100>=w.

AC通过的C++语言如下

 

 

### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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