// yukicoder: No.831 都市めぐり
// 2019.7.27 bal4u
// a(n) = (n^3+3*n^2+5*n-6)/6 if n is even;
// a(n) = (n^3+3*n^2+5*n-3)/6 if n is odd.

#include <stdio.h>

typedef long long ll;

int main()
{
	int n;
	ll ans;
	
	scanf("%d", &n);
	if (n == 1) ans = 0;
	else{
		ans = (((ll)n+3)*n+5)*n;
		if (n & 1) ans = (ans-3) / 6;
		else       ans = (ans-6) / 6;
	}
	printf("%lld\n", ans);
	return 0;
}