Codeforces 4A. Watermelon

最新推荐文章于 2020-01-14 15:43:00 发布

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本文介绍了一个简单的编程问题——如何判断一个整数是否可以分解为两个偶数之和。这个问题源自 Codeforces 的一道练习题，通过分析题目背景及给出的示例，我们提供了一段简洁的 Java 代码来解决这个问题。

【题目来源】：Codeforces 4A.

A. Watermelon

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

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).

【题目大意】：输入一个数判断是否能变为两个偶数之和。

（除2以外的任意偶数都可以化成两个偶数之和）

【AC代码】：

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
    	int n=sc.nextInt();
    	if(n%2==0&&n!=2)
    		System.out.print("YES");
    	else System.out.print("NO");
	}
}