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 Python (3.4.3) (judge verdict: Wrong Answer)

N = int(input())

if N<=3 or N > 100:
    print('No')
elif N % 4 == 0:
    print('Yes')
elif N % 7 == 0:
    print('Yes')
elif (N - 91) % 4 == 0:
    print('Yes')
elif (N - 84) % 4 == 0:
    print('Yes')
elif (N - 77) % 4 == 0:
    print('Yes')
elif (N - 70) % 4 == 0:
    print('Yes')
elif (N - 63) % 4 == 0:
    print('Yes')
elif (N - 56) % 4 == 0:
    print('Yes')
elif (N - 49) % 4 == 0:
    print('Yes')
elif (N - 42) % 4 == 0:
    print('Yes')
elif (N - 35) % 4 == 0:
    print('Yes')
elif (N - 28) % 4 == 0:
    print('Yes')
elif (N - 21) % 4 == 0:
    print('Yes')
elif (N - 14) % 4 == 0:
    print('Yes')
elif (N - 7) % 4 == 0:
    print('Yes')
else:
    print('No')