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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<map>
#include<vector>
#include<algorithm>

#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FUNCTION__ << endl;
#define int long long

const long long INF = __LONG_LONG_MAX__ / 2;

using namespace std;
int a[200000];
bool visited[200000];

signed main(){
    int N;
    cin >> N;
    for (int i = 1; i< N+1 ; i++) visited[i] = false;
    for (int i = 1; i < N+1; i++){
      cin >> a[i];
    }

    int count = 0;
    int pos = 1;
    visited[1] = true;
    while(true){
      if (pos == 2) {
        cout << count;
        return 0;
      }
      pos = a[pos];
      if (visited[pos]) {
        cout << -1 ;
        return 0;
      }
      visited[pos] = true;
      count++;
    }
    return 0;
}