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 <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <list>
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define p(x) cout << x << endl;

#define MOD 1000000007
#define PI 3.14159265359

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll n;
    cin >> n;
    vector<ll> l;

    l.push_back(2);
    l.push_back(1);

    for(int i = 2; i <= n; i++){
        ll num = l[i - 2] + l[i - 1];
        l.push_back(num);
    }

    p(l[n]);
}