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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

process.stdin.resume();
process.stdin.setEncoding('utf8');

var lines = [];
var reader = require('readline').createInterface({
  input: process.stdin,
  output: process.stdout
});
reader.on('line', (line) => {
  lines.push(line);
});
reader.on('close', () => {
var array = lines[0];

var searchkeyence = array.indexOf("keyence") ;
if (searchkeyence !== -1) {
console.log("YES");
} else {

var searchkeyenc = array.indexOf("keyenc") ;
var searche = array.lastIndexOf("e") ;
if (searchkeyenc !== -1 && searche !== -1) {
console.log("YES");
} else {

var searchkeyen = array.indexOf("keyen") ;
var searchce = array.lastIndexOf("ce") ;
if (searchkeyen !== -1 && searchce !== -1) {
console.log("YES");
} else {

var searchkeye = array.indexOf("keye") ;
var searchnce = array.lastIndexOf("nce") ;
if (searchkeye !== -1 && searchnce !== -1) {
console.log("YES");
} else {

var searchkey = array.indexOf("key") ;
var searchence = array.lastIndexOf("ence") ;
if (searchkey !== -1 && searchence !==-1) {
console.log("YES");
} else {

var searchke = array.indexOf("ke") ;
var searchyence = array.lastIndexOf("yence") ;
if (searchke !== -1 && searchyence !== -1) {
console.log("YES");
} else {

var searchk = array.indexOf("k") ;
var searcheyence = array.lastIndexOf("eyence") ;
if (searchk !== -1 && searcheyence !== -1) {
console.log("YES");
} else {
console.log("NO");
}
}
}
}
}
}
}
});