The only line of the input contains one integer n (3 ≤ n ≤ 30) — the amount of successive cars of the same make.

Output one integer — the number of ways to fill the parking lot by cars of four makes using the described way.

Let's denote car makes in the following way: A — Aston Martin, B — Bentley, M — Mercedes-Maybach, Z — Zaporozhets. For
n = 3 there are the following appropriate ways to fill the parking lot: AAAB AAAM AAAZ ABBB AMMM AZZZ BBBA BBBM BBBZ BAAA BMMM BZZZ MMMA MMMB MMMZ MAAA MBBB MZZZ ZZZA ZZZB ZZZM ZAAA ZBBB ZMMM

Originally it was planned to grant sport cars of Ferrari, Lamborghini, Maserati and Bugatti makes but this idea was renounced because it is impossible to drive these cars having small road clearance on the worn-down roads of IT City.

2*n-2个停车位，4种类型的车，n辆同一类型的A车要停在一起，旁边也不能是同一类型的A，问有多少种方法

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;

int main()

{

	int n;

	while(scanf("%d",&n)!=EOF)

	{

		__int64 a=3*n-1;

		__int64 aa=n-3;

		__int64 ans=a*3*pow(4,aa);

		printf("%lld\n",ans);

	}

	return 0;

}