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<bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define pb push_back
#define F first
#define S second
#define pi pair<int,int>
#define all(v) v.begin(),v.end()
const int MOD = 1e9+7;
template <typename A>istream& operator>>(istream& input,vector<A>& x) {for(auto& i:x)input>>i;return input;}
template<typename A>ostream& operator<<(ostream& output,vector<A>& x) {for(auto& i:x)output<<i<<' ';return output;}
signed main()
{
    ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
    int T=1;
   // cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        vi a(n+1);
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        int i=1;
        int cnt=1;
        while(a[i]!=2)
        {
            cnt++;
            i=a[i];
            if(cnt>2e5)
            {
                cout<<-1;
                return 0;
            }
        }
        cout<<cnt<<"\n";

    }

}