Problem:
Score : 200 points

Problem Statement

La Confiserie d'ABC sells cakes at 4 dollars each and doughnuts at 7 dollars each.
Determine if there is a way to buy some of them for exactly N dollars. You can buy two or more doughnuts and two or more cakes, and you can also choose to buy zero doughnuts or zero cakes.

Constraints

N is an integer between 1 and 100, inclusive.

Input

Input is given from Standard Input in the following format:

N

Output

If there is a way to buy some cakes and some doughnuts for exactly N dollars, print Yes; otherwise, print No.

Sample Input 1

11

Sample Output 1

Yes

If you buy one cake and one doughnut, the total will be 4 + 7 = 11 dollars.

Sample Input 2

40

Sample Output 2

Yes

If you buy ten cakes, the total will be 4 \times 10 = 40 dollars.

Sample Input 3

3

Sample Output 3

No

The prices of cakes (4 dollars) and doughnuts (7 dollars) are both higher than 3 dollars, so there is no such way.

Solution in Rust (1.15.1) (judge verdict: Accepted)

mod io{
    use std::io::*;
    pub fn read<T: ::std::str::FromStr>()->T{
        let stdin=stdin();
        let ret=stdin.lock().bytes().map(|c|c.unwrap()as char)
        .skip_while(|c|c.is_whitespace())
        .take_while(|c|!c.is_whitespace())
        .collect::<String>().parse().ok().unwrap();
        ret
    }
}
use io::read;
fn main(){
    let n=read::<i32>();
    let mut f=0;
    for i in 0..n/4+1{
        if (n-i*4)%7<1{f+=1}
    }
    println!("{}",if f>0{"Yes"}else{"No"})
}