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: Runtime Error)

#include<bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < n ; i++)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define whole(f, x, ...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x) // decltypeで型取得、引数があればva_argsのところに入れる
using namespace std;
typedef long long ll; // long longをllでかけるようにした
const int INF = 1e9;

int main(void){
    int n; cin >> n;
    vector<ll> A(n);
    A[0] = 2;
    A[1] = 1;
    FOR(i, 2, n + 1){
        A[i] = A[i - 1] + A[i - 2];
    }
    cout << A[n] << endl;

}