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 Python (3.4.3) (judge verdict: Accepted)

s = list(input())

if s[0] == "k":
    li = []
    for i in range(6):
        li.append(s[len(s)-(6-i)])
    if "".join(li) == "eyence":
        print("YES")
        quit()

if s[0] + s[1] == "ke":
    li = []
    for i in range(5):
        li.append(s[len(s)-(5-i)])
    if "".join(li) == "yence":
        print("YES")
        quit()

if s[0] + s[1] + s[2] == "key":
    li = []
    for i in range(4):
        li.append(s[len(s)-(4-i)])
    if "".join(li) == "ence":
        print("YES")
        quit()

if s[0] + s[1] + s[2] + s[3] == "keye":
    li = []
    for i in range(3):
        li.append(s[len(s)-(3-i)])
    if "".join(li) == "nce":
        print("YES")
        quit()

if s[0] + s[1] + s[2] + s[3] + s[4] == "keyen":
    li = []
    for i in range(2):
        li.append(s[len(s)-(2-i)])
    if "".join(li) == "ce":
        print("YES")
        quit()

if s[0]  + s[1] + s[2] + s[3] + s[4] + s[5] == "keyenc":
    if s[len(s) - 1] == "e":
        print("YES")
        quit()

        
lis = []
for i in range(7):
    lis.append(s[len(s)-i-1])
if "".join(lis) == "keyence":
    print("YES")
    quit()

print("NO")