Problem:
Score : 200 points

Problem Statement

It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers.

You are given an integer N. Find the N-th Lucas number.

Here, the i-th Lucas number L_i is defined as follows:

L_0=2

L_1=1

L_i=L_{i-1}+L_{i-2} (i≥2)

Constraints

1≤N≤86

It is guaranteed that the answer is less than 10^{18}.

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the N-th Lucas number.

Sample Input 1

5

Sample Output 1

11

L_0=2

L_1=1

L_2=L_0+L_1=3

L_3=L_1+L_2=4

L_4=L_2+L_3=7

L_5=L_3+L_4=11

Thus, the 5-th Lucas number is 11.

Sample Input 2

86

Sample Output 2

939587134549734843

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>
 
int main(void){
	
	double n;
	int l1,l2,l3,i;
	
	scanf("%lf",&n);
	
	if(n<1 || n>86 || ceil(n) != floor(n)){
		
	}else{
		l1 = 2;
		l2 = 1;
		for(i=1;i<n;i++){
			l3 = l1 + l2;
			l1 = l2;
			l2 = l3;
		}
		if(l3 < pow(10,18)){
			printf("%d\n",l3);
		}else{
		
		}
	}
	
	return 0;
}