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: Wrong Answer)

#include <bits/stdc++.h>
#include <math.h>
using namespace std;
int samecount(string S){
    int A = 0;
    for(int i = 0;i < S.size();i++){
        int QA = 1;
        for(int j = i ;j < S.size() - 1;j++){
           if(S.at(j) == S.at(j + 1)){
               QA++ ;
           }
           if(S.at(j) != S.at(j + 1)){
               break;
           }
           
        }
        A = max(QA,A);
    }
    return A;
}

int64_t minn(int64_t a,int64_t b){
    if(a >= b) return b;
    else return a;
}
int main(){
    int N;
    cin >>N;
    vector<int>  A(N);
    for(int i = 0;i <N;i++){
        cin >> A[i];
    }
   bool frag = 0;
   int memo = 1;
   int ans = 0;
   for(int i = 0;i < N;i++){
       memo = A[memo-1];
       ans++; cout << memo << endl;
       if(memo == 2){
           frag = 1; break; 
       }
   }
   if(frag)cout << ans << endl;
   else cout << -1 << endl;
   
}