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>
#define ll long long

int take()
{
    int n;
    scanf("%d",&n);
    return n;

}

ll takes()
{
    ll n;
    scanf("%lld",&n);
    return n;

}
#define M 200005
#define Int take()
#define LL takes()

#define fr(n) for(int i=1;i<=n;i++)
#define rfr(n) for(int i=n;i>=1;i--)
int cas;
#define print printf("Case %d:",++cas)
#define vi vector<int>
#define vl vector<ll>
#define pb push_back
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define mk  make_pair
#define fill(x,v) memset(x,v,sizeof x)
using namespace std;

ll dp[100];

int main(){
    ll n=LL;
    dp[0]=2;
    dp[1]=1;
    int pr=1;
    for(int i=2;i<=n;i++){
        dp[i]=dp[i-1]+dp[i-2];
    }
    cout<<dp[n]<<endl;
    return 0;
}