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: Wrong Answer)

import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;

public class Main {

	private final static long MOD = 1000000007;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		boolean ok = false;
		String keyence = "keyence";
		int index = 0;

		for (int i = 0; i < S.length(); i++) {
			if (S.charAt(i) == 'k') {
				index = 1;
			} else if (S.charAt(i) == keyence.charAt(index)) {
				index++;

				if (keyence.length() == index) {
					ok = true;
					break;
				}
			} else {
				if (S.indexOf(keyence.substring(index), i + 1) != -1) {
					ok = true;
					break;
				}
			}
		}

		System.out.println(ok ? "YES" : "NO");
	}

	private static boolean isDebug = System.getProperty("sun.desktop") != null;

	private static void debug(Object... o) {
		if (isDebug) {
			System.err.println(Arrays.deepToString(o));
		}
	}

	public static class Scanner {
		private BufferedInputStream inputStream;
		private StringBuilder buffer = new StringBuilder();

		public Scanner(InputStream in) {
			inputStream = new BufferedInputStream(in);
		}

		public int nextInt() throws IOException {
			int num = 0;
			int sign = 1;

			int read = skip();
			if (read == '-') {
				sign = -1;
				read = inputStream.read();
			}

			do {
				num = num * 10 + sign * (read - 0x30);
			} while ((read = inputStream.read()) > 0x20);

			return num;
		}

		public void fill(int[] a) throws IOException {
			for (int i = 0; i < a.length; i++) {
				a[i] = nextInt();
			}
		}

		public void fill(int[] a, int[] b) throws IOException {
			if (a.length != b.length) {
				throw new IllegalArgumentException();
			}

			for (int i = 0; i < a.length; i++) {
				a[i] = nextInt();
				b[i] = nextInt();
			}
		}

		public long nextLong() throws IOException {
			long num = 0;
			int sign = 1;

			int read = skip();
			if (read == '-') {
				sign = -1;
				read = inputStream.read();
			}

			do {
				num = num * 10 + sign * (read - 0x30);
			} while ((read = inputStream.read()) > 0x20);

			return num;
		}

		public void fill(long[] a) throws IOException {
			for (int i = 0; i < a.length; i++) {
				a[i] = nextLong();
			}
		}

		public void fill(long[] a, long[] b) throws IOException {
			if (a.length != b.length) {
				throw new IllegalArgumentException();
			}

			for (int i = 0; i < a.length; i++) {
				a[i] = nextLong();
				b[i] = nextLong();
			}
		}

		public long[] nextLong(int n) throws IOException {
			long[] array = new long[n];
			for (int i = 0; i < n; i++) {
				array[i] = nextLong();
			}

			return array;
		}

		public String next() throws IOException {
			int read = skip();
			do {
				buffer.append((char) read);
			} while ((read = inputStream.read()) > 0x20);

			String text = buffer.toString();
			buffer.delete(0, buffer.length());

			return text;
		}

		private int skip() throws IOException {
			int read;
			while ((read = inputStream.read()) <= 0x20)
				;

			return read;
		}
	}
}