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: Runtime Error)

#include<bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;++i)
#define rrep(i,n) for(int i=n-1;i>=0;--i)
#define yesno(flg) if(flg){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}
#define MAX_N 1002
#define i197 1000000007

using namespace std;

typedef long long ll;
typedef pair<ll,ll> P1;
typedef pair<int,int> Pi;
typedef pair<double,Pi> Pdi;
typedef pair<ll,int>Pli;
typedef pair<P1,ll> P2;
const ll INF=100000000000000001;
struct edge{int to,cost;};
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
struct Road{double cost;int a,b;};


int main() {

	ll n;
	ll a[10001];
	ll total=0;

	cin>>n;
	for(ll i=1; i<n+1; i++){
		cin>>a[i];
	}
	int i=1;
	int o=0;
	for(ll j=1; j<n+1; j++){
		if(a[i]==2){
			total++;
			break;
		}else{
			i=a[i];
			total++;
		}
		o++;
	}
	if(o==n){
		cout<<"-1"<<endl;
	}else{
		cout<<total<<endl;
	}
	return 0;
}