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 Ruby (2.3.3) (judge verdict: Wrong Answer)

cake = 4
donuts = 7
money = gets.to_i

cakeNum = 0
donutsNum = 0

result = "No"

donutsArray = Array.new
cakeArray = Array.new

(money/ cake + 1).times do
    cakeArray.push(cake * (cakeNum))
    cakeNum += 1
end
(money / donuts + 1).times do
    donutsArray.push(donuts * (donutsNum))
    donutsNum += 1
end

cakeArray.each do |cakeItem|
    if cakeItem == money then
        result = "Yes"
    end
    donutsArray.each do |donutsItem|
        if cakeItem + donutsItem == money || donutsItem == money then
            resutl = "Yes"
        end
    end
end
puts result