output_description: Print the N-th Lucas number.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
attempt: fn main() {
      let n: i32 = read();
      let ans = lucas(n);
      println!("{}", ans);
  }
  
  fn lucas(n: i32) -> i32 {
      if n == 0 { 2 }
      else if n == 1 { 1 }
      else {
          lucas(n-1) + lucas(n-2)
      }
  }
  
  // 以下関数
  
  use std::io::*;
  use std::str::FromStr;
  pub fn read<T: FromStr>() -> T {
      let stdin = stdin();
      let stdin = stdin.lock();
      let token: String = stdin
          .bytes()
          .map(|c| c.expect("failed to read char") as char)
          .skip_while(|c| c.is_whitespace())
          .take_while(|c| !c.is_whitespace())
          .collect();
      token.parse().ok().expect("failed to parse token")
  }
  
  pub const MOD: u64 = 10e9 as u64 + 7;
  
  pub fn is_prime(n: i32) -> bool {
      if n < 2 {
          return false;
      } else if n == 2 {
          return true;
      }
      let mut i = 2;
      while i * i < n {
          if n % i == 0 {
              return false;
          } else {
              i += 1;
              continue;
          }
      }
      true
  }
  
  pub fn lcm(mut n: i32, mut m: i32) -> i32 {
      n = if n > m { m } else { n };
      m = if n > m { n } else { m };
      let mut i = n;
      while i % m != 0 {
          i += n;
      }
      i
  }
  
  pub fn abs(n: i32) -> i32 {
      if n >= 0 {
          n
      } else {
          -n
      }
  }
  
  pub fn max(n: i32, m: i32) -> i32 {
      if n >= m {
          n
      } else {
          m
      }
  }
  
  pub fn min(n: i32, m: i32) -> i32 {
      if n <= m {
          n
      } else {
          m
      }
  }
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"}]