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 <algorithm>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <tuple>
#include <vector>

using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define reps(i, f, n) for (int i = (f); i < (n); i++)
ll const INF = 1LL << 60;

vector<ll> memo(100, -1);

ll lucas(ll n) {
    if (memo[n] != -1)
        return memo[n];

    return memo[n] = lucas(n-1) + lucas(n-2);
}

int main() {
    int N;
    cin >> N;

    memo[0] = 2;
    memo[1] = 1;

    cout << lucas(N) << endl;

   return 0;
}