1021: Fibonacci Again

Problem Description：

There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

Input

Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

Output

Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

Sample Input：

0

1

2

3

4

5

Sample Output：

no

no

yes

no

no

no

//  i    i=0 i=1 i=2 i=3   i=4 i=5 i=6 i=7   i=8 i=9 i=10 1=11
//f(i)%3  1   2   0   2     2   1   0   1     1   2   0    2   (按规律循环出现：2101 1202/ 2101 1202/...... )
//        no  no yes  no    no  no yes  no    no  no  yes
//由上规律可知yes出现位置为2 6 10 14 ......即i%4==2处！！！
#include <iostream>
using namespace std;

int main()
{
int i;
while(cin>>i)
{
if(i%4==2)
cout<<"yes"<<endl;
else
cout<<"no"<<endl;
}
return 0;
}