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"iomanip"
#include"iostream"
#include"limits"
#include"cmath"
#include"vector"
#include"algorithm"
#include"list"
#include"queue"
#include"stack"
#include"set"
#include"unordered_set"
#include"map"
#include"unordered_map"
#include"string"
#include"cstring"
#include"assert.h"
using namespace std;
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define mp make_pair
#define pb push_back
#define mset(a,b) memset(a,b,sizeof(a))
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	ll x=2,y=1;
	int n;
	cin>>n;
	for(int i=0;i<n-1;++i)
	{
		ll t=y;
		y=y+x;
		x=t;
	}
	cout<<y<<'\n';
}