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)

use std::io::BufReader;
use std::io::BufRead;
use std::io;

fn main() {
    let stdin = io::stdin();
    let mut stdin = BufReader::new(stdin.lock());
    let mut buf = String::new();
    let mut ans = "No";

    let _ = stdin.read_line(&mut buf);
    let n = buf.trim().parse::<i32>().unwrap();

    let max = n / 4 + 1;


    for i in 0..max {
        for j in 0..max {
            if n == i * 4 + j * 7 {
                ans = "Yes";
            }
        }
    }

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