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;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> l_l;
typedef vector<ll> vel;
typedef vector<int> vei;
typedef vector<string> ves;
typedef vector<vector<int>> ve_vei;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<(int)(n);i++)
#define rep2(i,n) for(int i=2;i<(int)(n);i++)
#define fs first
#define sc second
#define all(a) a.begin(),a.end()
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
 
int main() {
	//cout.precision(10);
	cin.tie(0);
	ios::sync_with_stdio(false);

    int N; cin >> N;
    vector<bool> check(N);
    vei a(N); 
    rep(i,N) {
        cin >> a[i];
        a[i]--;
        check[i] = false;
    }
    check[0] = true;

    int i = 0;
    int cnt = 1;
    while(cnt < N) {
        if(check[ a[i] ] == true){
            cout << -1 << endl;
            return 0;
        }
        else {
            if(a[i] == 1){
                cout << cnt << endl;
                return 0;
            }
            check[ a[i] ] = true;
            i = a[i];
            cnt++;
        }
    }
}