Problem:
Score : 100 points

Problem Statement

Shichi-Go-San (literally "Seven-Five-Three") is a traditional event in a certain country to celebrate the growth of seven-, five- and three-year-old children.

Takahashi is now X years old. Will his growth be celebrated in Shichi-Go-San this time?

Constraints

1 ≤ X ≤ 9

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

If Takahashi's growth will be celebrated, print YES; if it will not, print NO.

Sample Input 1

5

Sample Output 1

YES

The growth of a five-year-old child will be celebrated.

Sample Input 2

6

Sample Output 2

NO

See you next year.

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Numerics;
using System.Linq;
using System.Collections.Generic;
using System.Text;
using System.Collections;

namespace debug
{
    class main
    {
        static void Main(string[] args)
        {
            //問題クラスを展開
            ProgramA a = new ProgramA();
            a.main();//実行する

        }

    }

    class ProgramA
    {
        public void main()
        {
            //入力
            int n = int.Parse(Console.ReadLine());

            if (n == 3 || n == 5 || n == 7)
                Console.WriteLine("YES");
            else
                Console.WriteLine("NO");
        }
    }

    class ProgramB
    {
        public void main()
        {

            //入力
            string[] s = Console.ReadLine().Split(' ');
            int n = int.Parse(s[0]);
            int m = int.Parse(s[1]);
            int c = int.Parse(s[2]);

            int count = 0;

            string[] b = Console.ReadLine().Split(' ');
            for (int i = 0; i < n; i++)
            {
                string[] a = Console.ReadLine().Split(' ');
                int sum = c;
                for (int j = 0; j < m; j++)
                    sum += int.Parse(a[j]) * int.Parse(b[j]);

                if (sum > 0)
                    count++;
            }



            //答え出力
            Console.WriteLine(count);


        }
    }

    class ProgramC
    {
        public void main()
        {

            //入力
            string[] ss = Console.ReadLine().Split(' ');
            int n = int.Parse(ss[0]);
            long m = long.Parse(ss[1]);

            //i文字目までにACが何個あるか記録する
            int[] AC = new int[n + 1];
            int count = 0;

            List<Task> list = new List<Task>();

            for (int i = 0; i < n; i++)
            {
                string[] s = Console.ReadLine().Split(' ');
                list.Add(new Task(long.Parse(s[0]), long.Parse(s[1])));
            }

            var sort = list.OrderBy(x => x.a);
            long ans = 0;

            foreach (var s in sort)
            {
                if (m > s.b)
                {
                    ans += s.a * s.b;
                    m -= s.b;
                }
                else
                {
                    ans += s.a * m;
                    break;
                }

            }

            Console.WriteLine(ans);

        }


    }

    class Task
    {
        public long a;
        public long b;
        public Task(long a, long b)
        {
            this.a = a;
            this.b = b;
        }

    }




    class ProgramD
    {
        public void main()
        {

            //入力
            string[] s = Console.ReadLine().Split(' ');
            long a = long.Parse(s[0]);
            long b = long.Parse(s[1]);


            //F(A,B) は F(0,A-1)とF(0,B)との排他的論理和（よってF(0,n)を考える）
            //排他的論理和はどっからでも足してよい。また任意の偶数nにおいてnとn+1の排他的論理和は1である。

            long ta = 0;
            long tb = 0;
            //a-1を出す
            if (a == 0)
            {

            }
            else if ((a - 1) % 2 == 1)
            {
                if ((a / 2) % 2 == 1)
                    ta = 1;
            }
            else
            {
                if (((a - 1) / 2) % 2 == 1)
                    ta = a;
                else
                    ta = a - 1;
            }

            //bを出す
            if (b % 2 == 1)
            {
                if (((b + 1) / 2) % 2 == 1)
                    tb = 1;
            }
            else
            {
                if ((b / 2) % 2 == 1)
                    tb = b + 1;
                else
                    tb = b;
            }

            //taとtbの排他的論理和を求める
            long ans = (ta ^ tb);

            //答え出力
            Console.WriteLine(ans);

        }

    }


}