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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define long long long
#define up(i,a,b) for (int i=a; i<=b; i++)
#define down(i,a,b) for (int i=a; i>=b; i--)
#define endl '\n'
#define X first
#define Y second
#define II pair<int, int>
#define III pair<int, pair<int, int> >
#define debug(X) cerr<< #X << " = " <<X << endl
#define debug2(X,Y) cerr<< #X << " = " <<X << ","<<#Y<<" = "<<Y<<endl
#define show(X,a,b) {cerr << #X << " = "; up(__,a,b) cerr << X[__] << ' '; cerr << endl;}
#define gc getchar
#define pc putchar
using namespace std;
int n;
long a[100];
void input()
{
    cin>> n;
    a[0]= 2; a[1]= 1;
    up(i,2,n) a[i]= a[i-1]+ a[i-2];
    cout<<a[n];
}
void solve()
{

}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);// don't use when interactive

    #ifdef I_Love_Pork
    #define TASK "tmp"
    freopen(TASK".inp","r",stdin);
    freopen(TASK".out","w",stdout);
	  #endif

    input();
    solve();

    return 0;
}