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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
 
public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        String[] arr = br.readLine().split("");
        boolean res = false;
        boolean isRemoving = false;
        boolean isFirstRemoving = true;
        int index = 0;
        String[] keyence = {"k", "e", "y", "e", "n", "c", "e"};

        for (int i = 0; i < arr.length; i++) {
            String s = arr[i];
            if (isRemoving) {
                if (s.equals(keyence[index])) {
                    index++;
                    isRemoving = false;
                } 
            } else {
                if (s.equals(keyence[index])) {
                    index++;
                    if (index == 7) {
                        if (isFirstRemoving || i == arr.length - 1) {
                            res = true;
                            break;
                        } else {
                            res = false;
                            break;
                        }
                    }
                } else {
                    if (isFirstRemoving) {
                        isRemoving = true;
                        isFirstRemoving = false;
                    } else {
                        res = false;
                        break;
                    }
                }
            }
        }
        out.println(res ? "YES" : "NO");
        out.close();
    }
}