// The first problem is too easy to write its solution
// The second problem of the test held on 13 November was taken from
// the assignment 7 which was held on Wednesday.
// The solution of the third problem is provided below.
// It is not well documented. Please read the file analysis.pdf
// for its underlying logic.
// The test cases and grading policy for the test are mentioned in the
// file   test_cases_Thu.pdf (gradingpolicy is on the last page).

class right_tilings
{
    public static int M(int n)
    {
        if(n==0) return 1;
        else return (E(n) + N(n-1));
    }

    public static int N(int n)
    {
        if(n==0) return 1;
        else 
	{
            if(n==1) return 2;
            else return (C(n) + O(n-1));

	}    
    }

    public static int O(int n)
    {
        if(n==1) return 1;
	else return (M(n-1)+O(n-1));
    }

    public static int C(int n)
    {
	if(n==0) return 1;
	else
	{
	    if(n==1) return 2;
	    else return (D(n-1) + 2*D(n-2) + C(n-2) + M(n-1));
	}
    }

    public static int D(int n)
    {
	if(n==0) return 1;
	else
	{
	    if(n==1) return 2;
	    else return (D(n-1) + 2*D(n-2) + 2*C(n-2) + E(n-1));
	}
    }

    public static int E(int n)
    {
	if(n==0) return 1;
	else
	{
	    if(n==1) return 2;
	    else return (D(n-1) + 2*D(n-2) + 2*N(n-2) + E(n-1));
	}
    }
    public static void main(String args[])
    {   
	int n = Integer.parseInt(args[0]);
	int total_count = D(n);
	System.out.println("Total number of possible tilings= "+total_count);
    }
}