[A - Problem A]

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

即给出一个数，能不能将其分为两个偶数。

程序分析

若一个数能被被拆分为两个偶数的和，则该数必为偶数，因此程序需判定输入的数是否为偶数即可。此外需认识到“2”也为偶数，但它并不符合要求。

AC程序如下：

//CodeForces-4A
#include<iostream>
using namespace std;
int main()
{
    int w;
    while (cin >> w)
    {
        if (w <= 2)
            cout << "no";
        else
        {
            if (w % 2 == 0)
                cout << "yes";
            else cout << "no";
        }
    }
    return 0;
}