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 <stdlib.h>
#include <string.h>

int lucas_N =0;

void lucas(int L0, int L1, int n)
{
  if(n != 1){
    lucas_N = L0 + L1;
    lucas(L1, lucas_N , n-1); 
  }
}

int main()
{

    int  N;
 // 整数の入力
    scanf("%d", &N);
  
    if(N>=2){
      lucas(2,1,N);
    } else {
      lucas_N =1;
    }

printf("%d\n", lucas_N);

    return 0;
}