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: Accepted)

#include <stdio.h>

int main(){
 int N, n, i;
 unsigned long a[3];
 a[0] = 2;
 a[1] = 1;

 scanf("%d",&N);
 for(i=0; i<N-1;i++){
  n = i % 3;
  a[(n+2)%3] = a[n] + a[(n+1)%3];
  //printf("%3d:%ld\n",i+3,a[(n+2)%3]);
 }
 if(N==1){
  printf("1");
 }else if(N==0){
  printf("2");
 }else{
  printf("%ld",a[(n+2)%3]);
 }
 return 0;
}