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 <algorithm>
#include <cstring>
#include <vector>
#include <numeric>
#include <cstdlib>
#include <queue>
#include <stack>
#include <climits>
#include <array>
using namespace std;

typedef long long ll;
#define pb push_back

#define rep(i, n) for (int i=0;i<(n);++i)
#define rep2(i, s, n) for (int i=s;i<(n);++i)
#define rev(i, n) for (int i=(n);i>0;--i)
#define INF (1e9+1e9)


ll l (array<ll, 1000>& L, int n) {
    if (L[n] != -1) return L[n];
    if (n == 0) return 2;
    if (n == 1) return 1;
    L[n] = l(L,n-1) + l(L,n-2);
    return L[n];
}

int main() {
    int n; cin>>n;
    array<ll, 1000> L;
    L.fill(-1);    

    cout << l(L,n) << endl;
}