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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define repI(i, d, n) for (int i = (d); i < (n); ++i)
#define reps1(i, n) for (int i = 1; i < (n); ++i)
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using ll = long long;
using P = pair<int, int>;
using Pl = pair<ll, ll>;
using M = map<int, int>;
using Ml = map<ll, ll>;
ll const INF = 1ll<<61;
double pi = 3.141592653589793238;

int main() {
  ll n;
  cin >> n;

  vector<ll> l(n+1);
  l[0] = 2;
  l[1] = 1;
  repI(i, 2, n+1) l[i] = l[i-1] + l[i-2];

  cout << l[n] << endl;

  return 0;
}