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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <utility>

#include <cassert>
#include <cmath>
#include <cstdio>

using namespace std;

#define fi first
#define se second
#define mkp make_pair
#define pb push_back
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define FOR(i,a,b) for(ll i=(a); i < (b); i++)
#define RFOR(i,a,b) for(ll i=(b)-1; i >= (a); i--)
#define rep(i,n) FOR(i,0,(n))
#define rrep(i,n) RFOR(i,0,(n))

int main() {
	string K = "keyence";
	string S;
	cin >> S;
	bool flg = false;
	for(int i = 0; i < K.size(); i++) {
		int a = S.find(string(K.begin(), K.begin()+i));
		if(a != string::npos && S.find(string(K.begin()+i, K.end()), a+i) != string::npos)
			flg = true;
	}
	if(flg)
		cout << "YES" << endl;
	else
		cout << "NO" << endl;
	return 0;
}