output_description: Print the N-th Lucas number.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
attempt: import sys
  
  sys.setrecursionlimit(10**8)
  
  
  def ii():
      return int(sys.stdin.readline())
  
  
  def mi():
      return map(int, sys.stdin.readline().split())
  
  
  def li():
      return list(map(int, sys.stdin.readline().split()))
  
  
  def li2(N):
      return [list(map(int, sys.stdin.readline().split())) for _ in range(N)]
  
  
  def dp2(ini, i, j):
      return [[ini] * i for _ in range(j)]
  
  
  def dp3(ini, i, j, k):
      return [[[ini] * i for _ in range(j)] for _ in range(k)]
  
  
  # import bisect #bisect.bisect_left(B, a)
  # from collections import defaultdict #d = defaultdict(int) d[key] += value
  # from itertools import accumulate #list(accumulate(A))
  
  N = ii()
  
  A = [0] * (N + 1)
  
  A[0] = 2
  A[1] = 1
  
  for i in range(2, N + 1):
      A[i] = A[i - 1] + A[i - 2]
  
  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"}]