4A - Watermelon

最新推荐文章于 2024-08-13 08:00:00 发布

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最新推荐文章于 2024-08-13 08:00:00 发布

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分类专栏： =====codeforces=====

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本文介绍了一个简单的编程问题——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。通过分析西瓜重量的特性，给出了一个快速判断的方法，并提供了C++实现代码。

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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; andNO 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).

取决于奇数或者偶数，但是注意特判2的时候....

#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		if(n&1||n==2)
		{
			printf("NO\n");
		}
		else
		{
			printf("YES\n");
		}
	}
	return 0;
}