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;
// #define rep(i, c) for (int i = 0; i < (int)c; i++)
// int main()
// {
//   int N;
//   cin >> N;
//   if (N = 0)
//   {
//     cout << 2 << endl;
//   }
//   if (N = 1)
//   {
//     cout << 1 << endl;
//   }
//   int ans = 0;
//   vector<int> L;
//   for (int i = 2; i <= N; i++)
//   {
//     ans = ans + L.at(i)
//   }
// }

#include <bits/stdc++.h>
using namespace std;
typedef long long ll; //long long型に別名を付けてる
int N;
ll L[87]; //long long型のL。最大87要素。
int main()
{
  scanf("%d", &N);
  L[0] = 2LL, L[1] = 1LL; //int64_t型として扱うためにはLLと書く
  for (int i = 2; i <= N; i++)
    L[i] = L[i - 2] + L[i - 1]; //L.at(i)の数列
  printf("%lld \n", L[N]);   //L.at(N)をlong long型の数字＋改行で出力
}