Problem:
Score : 200 points

Problem Statement

Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.

The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up.
These buttons are numbered 1 through N.
When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i.
When Button i is not lighten up, nothing will happen by pressing it.

Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.

Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.

Constraints

2 ≤ N ≤ 10^5

1 ≤ a_i ≤ N

Input

Input is given from Standard Input in the following format:

N
a_1
a_2
:
a_N

Output

Print -1 if it is impossible to lighten up Button 2.
Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.

Sample Input 1

3
3
1
2

Sample Output 1

2

Press Button 1, then Button 3.

Sample Input 2

4
3
4
1
2

Sample Output 2

-1

Pressing Button 1 lightens up Button 3, and vice versa, so Button 2 will never be lighten up.

Sample Input 3

5
3
3
4
2
4

Sample Output 3

3

Solution in Haskell (GHC 7.10.3) (judge verdict: Time Limit Exceeded)

import Control.Monad                                                                                                       
                                                                                                                           
main :: IO ()                                                                                                              
main = do                                                                                                                  
  n <- read <$> getLine                                                                                                    
  as <- replicateM n (read <$> getLine)                                                                                    
                                                                                                                           
  print $ f n as 0 1                                                                                                       
                                                                                                                           
f :: Int -> [Int] -> Int -> Int -> Int                                                                                     
f n xs m i                                                                                                                 
  | i == 2 = m                                                                                                             
  | n == m = -1                                                                                                            
  | otherwise = f n xs (m+1) (xs!!(i-1))