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++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <fstream>
#include <string>
#include <math.h>
#include <cstdlib>
#include <istream>
#include <sstream>
#include <cctype>
#include <iomanip>
//cout << fixed << setprecision(20) << *** << endl;
#define rep(i,N) for(int i=0;i<(int)N;++i)
typedef long long ll;
using namespace std;
int main()
{
	string s;
	ll a,b;
	cin>>s;
	a=s.find("keyence");
	if(a!=string::npos){
		cout<<"YES"<<endl;
	}
	else{
		a=s.find("k");
		b=s.rfind("eyence");
		cout<<a<<":"<<b<<endl;
		if(a!=string::npos&&b!=string::npos&&a+1<b){
			cout<<"YES"<<endl;
			return 0;
		}
		a=s.find("ke");
		b=s.rfind("yence");
		cout<<a<<":"<<b<<endl;
		if(a!=string::npos&&b!=string::npos&&a+2<b){
			cout<<"YES"<<endl;
			return 0;
		}
		a=s.find("key");
		b=s.rfind("ence");
		cout<<a<<":"<<b<<endl;
		if(a!=string::npos&&b!=string::npos&&a+3<b){
			cout<<"YES"<<endl;
			return 0;
		}
		a=s.find("keye");
		b=s.rfind("nce");
		cout<<a<<":"<<b<<endl;
		if(a!=string::npos&&b!=string::npos&&a+4<b){
			cout<<"YES"<<endl;
			return 0;
		}
		a=s.find("keyen");
		b=s.rfind("ce");
		cout<<a<<":"<<b<<endl;
		if(a!=string::npos&&b!=string::npos&&a+5<b){
			cout<<"YES"<<endl;
			return 0;
		}
		a=s.find("keyenc");
		b=s.rfind("e");
		cout<<a<<":"<<b<<endl;
		if(a!=string::npos&&b!=string::npos&&a+6<b){
			cout<<"YES"<<endl;
			return 0;
		}
		cout<<"NO"<<endl;
	}
}