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 Rust (1.15.1) (judge verdict: Accepted)

use std::str::FromStr;

fn get_one<T : FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().parse().ok().unwrap();
}

fn get_vec<T : FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().split_whitespace().map(|e| e.parse().ok().unwrap()).collect();
}

fn main() {
    let n : usize = get_one();
    let mut v : Vec<usize> = Vec::new();
    for _i in 0..n {
        let x : usize = get_one();
        v.push(x-1);
    }
    let mut visited = vec![false; n];
    let mut cur : usize = 0;
    let mut ans = false;
    let mut cnt = 0;
    loop {
        //println!("{}", cur);
        if visited[cur] {
            break;
        }
        if cur == 1 {
            ans = true;
            break;
        }
        cnt += 1;
        visited[cur] = true;
        cur = v[cur];
    }
    if ans {
        println!("{}", cnt);
    }
    else {
        println!("{}", -1);
    }
}