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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::prelude::*;
use std::io::stdin;
use std::str::FromStr;

fn main() {
    let s: String = read();

    let keyence = "keyence";
    let mut count = 0;

    for (c1, c2) in s.chars().zip(keyence.chars()) {
        if c1 == c2 {
            count += 1;
        } else {
            break;
        }
    }
    for (c1, c2) in s.chars().rev().zip(keyence.chars().rev()) {
        if c1 == c2 {
            count += 1;
        } else {
            break;
        }
    }

    let ans = if count >= keyence.len() { "YES" } else { "NO" };

    println!("{}", ans);
}

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|b| b.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}