output_description: Print the N-th Lucas number.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
attempt: from statistics import mean, median, variance, stdev
  import numpy as np
  import sys
  import math
  import fractions
  import itertools
  
  
  def j(q):
      if q == 1:
          print("Yes")
      else:
          print("No")
      exit(0)
  
  
  def ct(x, y):
      if x > y:
          print("")
      elif x < y:
          print("")
      else:
          print("")
  
  
  def ip():
      return int(input())
  
  
  def swap(s, pos):
      s[pos] -= 1
      s[pos + 1] += 1
      print(s)
      if pos == n - 2:
          return s
      if s[pos + 2] == 0:
          swap(s, pos + 1)
      return s
  
  
  n = ip()  # 入力整数1つ
  # left,n= (int(i) for i in input().split())      #入力整数横2つ
  # n,x,y = (int(i) for i in input().split())    #入力整数横3つ
  # n,m,x,y= (int(i) for i in input().split())  #入力整数横4つ
  # a = [int(i) for i in input().split()]        #入力整数配列
  # a = input()                                  #入力文字列
  # a = input().split()                          #入力文字配列
  # n = ip()                      #入力セット(整数改行あり)(1/2)
  # a=[ip() for i in range(n)]    #入力セット(整数改行あり)(2/2)
  
  
  # jの変数はしようできないので注意
  # 全足しにsum変数使用はsum関数使用できないので注意
  
  a = [2, 1]
  for i in range(n - 1):
      a.append(a[i] + a[i + 1])
  print(a[n])
  
problem_description: 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)
samples: [{"input": "5", "output": "11\n    \n\n  * L_0=2\n  * L_1=1\n  * L_2=L_0+L_1=3\n  * L_3=L_1+L_2=4\n  * L_4=L_2+L_3=7\n  * L_5=L_3+L_4=11\n\nThus, the 5-th Lucas number is 11.\n\n* * *"}, {"input": "86", "output": "939587134549734843"}]