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>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
constexpr int di[] = { 0, 1, 0, -1 }, dj[] = { 1, 0, -1, 0 };
#define rep(i, n) for (int i = 0; i < n; i++)
#define repr(i, n) for (int i = n; i >= 0; i--)
#define SORT(v) sort((v).begin(), (v).end())
#define SORTR(v) sort((v).rbegin(), (v).rend())
#define all(v) (v).begin(), (v).end()
#define vmax 10000
constexpr ll inf = 100000000000000;
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/

int main(){
  int n;
  cin>>n;
  ll dp[87];
  for(int i=0;i<87;i++){
    dp[i]=inf;
  }
  dp[0]=2;
  dp[1]=1;
  for(int i=2;i<=n;i++){
    dp[i]=dp[i-1]+dp[i-2];
  }
  cout<<dp[n]<<endl;
  return 0;
}