Problem:
Score : 100 points

Problem Statement

A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.

Given a string S, determine whether it is coffee-like.

Constraints

S is a string of length 6 consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S is coffee-like, print Yes; otherwise, print No.

Sample Input 1

sippuu

Sample Output 1

Yes

In sippuu, the 3-rd and 4-th characters are equal, and the 5-th and 6-th characters are also equal.

Sample Input 2

iphone

Sample Output 2

No

Sample Input 3

coffee

Sample Output 3

Yes

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

using System;
using System.Collections.Generic;

namespace cswork
{
    class Program
    {
        class Scanner
        {
            string[] s;
            int i;

            char[] cs = new char[] { ' ' };

            public Scanner()
            {
                s = new string[0];
                i = 0;
            }

            public string next()
            {
                if (i < s.Length) return s[i++];
                string st = Console.ReadLine();
                while (st == "") st = Console.ReadLine();

                s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
                i = 0;
                return next();
            }

            public int nextInt()
            {
                return int.Parse(next());
            }

            public long nextLong()
            {
                return long.Parse(next());
            }

            public double nextDouble()
            {
                return double.Parse(next());
            }
        }

        public Program() { }

        static void Main(string[] args)
        {
            new Program().Omame();
        }

        Scanner cin;
        const int MAXIN = 1024;
        bool[] used = new bool[MAXIN];
        int[] perm = new int[MAXIN];

        List<int[]> permList = new List<int[]>();

        public void Permutation(int pos, int n, int length)
        {
            if (pos == n)
            {
                permList.Add((int[])perm.Clone());
            }

            for (int i = 0; i < n; i++)
            {
                if (!used[i])
                {
                    perm[pos] = i;
                    used[i] = true;
                    Permutation(pos + 1, n, length);
                    used[i] = false;
                }
            }
        }

        void Omame()
        {
            cin = new Scanner();
            string N = cin.next();

            if(N[2] == N[3] && N[4] == N[5])
            {
                Console.WriteLine("Yes");
            }
            else
            {
                Console.WriteLine("No");
            }
        }

    }
}