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 fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

using namespace std;

typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll, ll> pll;
typedef long double ld;

const ll mod = (ll)1e9 + 7;
const ll mxN = 200200;



void solve(){
	ll n;
	cin>>n;
	if(n==0) cout<<2<<"\n";
	else if(n==1) cout<<1<<"\n";
	else{
		ll dp[n+1];
		for(int i=0;i<=n;i++)
			if(i<2)
				dp[i]=2-i;
			else 
				dp[i]=dp[i-1]+dp[i-2];
		cout<<dp[n]<<"\n";
	}




}


int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.precision(20);
	cout << fixed;

	//ll t;cin>>t;while(t--)
	solve();

}