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: Accepted)

#include <iostream>
#include <algorithm>
#include <bitset>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <cstring>
#include <utility>
#include <vector>
#include <complex>
#include <valarray>
#include <fstream>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <numeric>

#define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X))
#define rrep(X,Y) for (int (X) = Y-1;(X) >= 0;--(X))
#define repa(X,A,Y) for (int (X) = A;(X) <= (Y);++(X))
#define all(X) (X).begin(),(X).end()
#define mod(n, m) (((n)%(m)+(m))%m)
#define fi first
#define sc second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pii;
typedef pair<ll,ll> Pll;
const int INFINT = 1 << 30;                          // 1.07x10^ 9
const ll INFLL = 1LL << 60;                          // 1.15x10^18
const double EPS = 1e-10;
const int MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

string S;

int main() {
  std::ios::sync_with_stdio(false);std::cin.tie(0);

  cin >> S;

  if (S.substr(0, 7) == "keyence" || S.substr(S.length()-7) == "keyence") {
      cout << "YES" << endl;
      return 0;
  }

  for (int l = 0; l < S.length()-1; ++l) {
      for (int r = l+1; r < S.length(); ++r) {
          string tmp;
          tmp = S.substr(0, l+1);
          tmp += S.substr(r);
          if (tmp == "keyence") {
              cout << "YES" << endl;
              return 0;
          }
      }
  }

  cout << "NO" << endl;

  return 0;
}