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>

int main(void) {
	long long int luc[86];
	int N;
	int i;
	luc[0] = 2;
	luc[1] = 1;

	scanf("%d", &N);
	for(i = 2; i < 86; i++) {
		luc[i] = luc[i-1] + luc[i-2];
	}
	printf("%d", luc[N-1]);

	return 0;
}