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)

//OM
#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <utility>
#include <sstream>
#include <algorithm>
using  namespace  std;

#define  x first
#define  y second
#define  pb push_back
#define  mp make_pair
#define  PI (acos(-1.0))
#define  mem(a,b) memset(a,b,sizeof(a))
#define  Sort(x) sort(x.begin(),x.end())
#define  FOR(i, b, e) for(int i = b; i <= (int)(e); i++)
#define  FORR(i, b, e) for(int i = b; i >=(int)(e); i--)
#define  FORI(i, s) for (__typeof ((s).end ()) i = (s).begin (); i != (s).end (); ++i)
#define  pr(x) cout<<x<<"\n"
#define  prs(x) cout<<x<<" "
#define  pr2(x,y) cout<<x<<" "<<y<<"\n"
#define  pr3(x,y,z) cout<<x<<" "<<y<<" "<<z<<"\n"
#define  ppr(a) cout<<a.x<<" "<<a.y<<"\n"
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)

typedef  long long ll;
typedef  pair <int, int> pii;
typedef  pair <double, double> pdd;
typedef  vector <int> vi;
typedef  vector <pii> vpii;


//int dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction
//int dx[]={1,1,0,-1,-1,-1,0,1};
//int dy[]={0,1,1,1,0,-1,-1,-1};//8 Direction

#define  EPS 1e-9
#define  MAX 100007
#define  MOD 1000000007

int main()
{
//    READ("in.txt");
//    WRITE("out.txt");
    int T,n;
    ll a[100];
    scanf("%d",&n);
    a[0]=2;
    a[1]=1;
    FOR(i,2,n){
        a[i]=a[i-1]+a[i-2];
    }
    pr(a[n]);
    return 0;
}