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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define repr(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define invrepr(i,a,b) for(int i=b-1;i>=a;i--)
#define invrep(i,n) invrepr(i,0,n)
#define repitr(itr,a) for(auto itr=a.begin();itr!=a.end();++itr)
#define P pair<int,int>
const int MOD=998244353;
const int MAX=1e5+10;
const ll INF=1e18;
const double PI=acos(-1);


int main() {
    ios_base::sync_with_stdio(false);

    string s;
    cin >> s;
    int n=s.size();
    string t="keyence";
    bool flag=0;
    rep(i,7) {
        bool f=1;
        rep(j,i) {
            if (s[j]!=t[j]) f=0;
        }
        repr(j,i,7) {
            if (s[n-7+j]!=t[j]) f=0; 
        }
        if (f) flag=1;
    } 
    if (flag) cout << "YES" << endl;
    else cout << "NO" << endl;

    return 0;   
}