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<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iomanip>
typedef long long int ll;
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define REP(i,n) for (int i=0;i<(n);i++)
#define EREP(i,n) for (int i=1;i<=(n);i++)
#define ALL(a) (a).begin(),(a).end()
//#define EVEL 1

#ifdef EVEL
#define DEB(X) cout << #X << ":" <<X<<" " ;
#define TF(f) f ? cout<<"true  " : cout<<"false ";
#define END cout<<"\n";
#else
#define DEB(X) {}
#define TF(f) {}
#define END {}
#endif
const int MOD = 1000000007;
const int INF = 2000000000;

    ll N,ans;
int main(){
    ios_base::sync_with_stdio(false);
    cin>>N;
    ll a,b;
    if(N==0)ans=2;
    else if(N==1)ans=1;
    else{
        b=2,a=1;
        REP(i,N-1){
            ans=a+b;
            b=a;
            a=ans;
        }
    }
    cout<<ans;
    return 0;
}