Problem:
Score : 200 points

Problem Statement

A string is called a KEYENCE string when it can be changed to keyence by removing its contiguous substring (possibly empty) only once.

Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.

Constraints

The length of S is between 7 and 100 (inclusive).

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S is a KEYENCE string, print YES; otherwise, print NO.

Sample Input 1

keyofscience

Sample Output 1

YES

keyence is an abbreviation of key of science.

Sample Input 2

mpyszsbznf

Sample Output 2

NO

Sample Input 3

ashlfyha

Sample Output 3

NO

Sample Input 4

keyence

Sample Output 4

YES

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
using System.IO;
using static System.Console;


namespace atcorder2
{
    class Program
    {
        public static long Read()
        {

            return long.Parse(Console.ReadLine());
        }

        public static long[] Reads()
        {

            return Console.ReadLine().Split().Select(long.Parse).ToArray();
        }

       
         public static long GCM(long x,long y)
        {
            if (x < y)  GCM(y, x);


            while (y > 0)
            {
                var z = y;
                y = x % y;
                x = z;
            }
            
            return x;
        }

        



        static void Main(string[] args)
        {
            var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            Console.SetOut(sw);


            var s = ReadLine();
            
            var c = "keyence";
           
            
            bool f = false;

            for (int i = 0; i < s.Length-7; i++)
            {
                if (s.Contains(c)) f = true;
            }
            int j=0, k = 6;
            for ( j = 0; j < 7;j++)
            {
                if (s[j] != c[j]) break;           
            }
            for (int l = s.Length-1; l >=0&&k>=0; l--)
            {
                if (s[l] != c[k]) break;
                k--;
            }
            


            
            WriteLine(f||k-j<=1 ? "YES" : "NO");
            

            






            Console.Out.Flush();
        }
    }
}