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: Wrong Answer)

#include <bits/stdc++.h>
#include <stdio.h>
#include <math.h>
using namespace std;
 
#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(), (v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define BIT(x,n) bitset<n>(x)
 
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<int,P> PP;
 
struct edge {
	int to,cost;
	edge(int t,ll c):to(t),cost(c) {}
};
 
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
int ddx[]={1,1,1,0,-1,-1,-1,0},ddy[]={1,0,-1,-1,-1,0,1,1};
 
//-----------------------------------------------------------------------------
 
int n;
 
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
 
	cin>>n;
	if(n==1) cout<<2<<endl;
	else if(n==2) cout<<1<<endl;
	else {
		ll n1=2,n2=1,now;
		REP(i,n-1) {
			now=n1+n2;
			n1=n2,n2=now;
		}
		cout<<now<<endl;
	}
 
	return 0;
}