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: Wrong Answer)

use std::str::FromStr;
use std::fmt::Debug;

fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn main(){
    let line = read_as_string();
    let keyence = String::from("keyence");

    for i in 1..keyence.len(){
        let(first, last) = keyence.split_at(i);
        let first_pos = line.find(first);
        let last_pos = line.find(last);
        //println!("first = {}, last = {}", first, last);
        if let(Some(first_pos), Some(last_pos)) = (first_pos, last_pos){
            if first_pos + first.len() <= last_pos{
                let pdust = first_pos;
                let mdust = last_pos - first_pos - first.len();
                let sdust = line.len() - last_pos - last.len();
                //println!("prefix = {}, middle={}, suffix={}", pdust, mdust, sdust);
                if (pdust == 0 && mdust == 0) ||
                    (mdust == 0 && sdust == 0) ||
                    (pdust == 0 && sdust == 0){
                        println!("YES");
                        return;
                    }
            }
        }
    }
    println!("NO");
}