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 <cmath>
#include <numeric>
using namespace std;
using ll=long long;
#define rep(i,a,b) for(int i=(a); i<(b); ++i) // a ≦ i < b 
#define Rrep(i,a,b) for(int i=(a);i>=(b);--i) // reverse repeat. a から b まで減少．
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend() //逆イテレータ
#define RANGE(a,b,c) (a).begin()+b,(a).begin()+c // コンテナ a の 要素 b から c へのイテレータ
#define MOD 1000000007
// typedef
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef vector<PII> VP;


int main() {
int N; cin >> N;
VI a(N); rep(i,0,N) scanf("%d", &a[i]);

unordered_map<int, bool> app; // 押したボタンをtrue

app[1] = true;
int sub = a[0] - 1, cnt=1;
while(a[sub] != 2){
    if (app[a[sub]]) {
        cout << -1 << "\n";
        return 0;
    }
    app[a[sub]] = true;
    sub = a[sub] - 1;
    ++cnt;
}
++cnt;
cout << cnt << "\n";


return 0;
}

// 境界，出力文字列 チェック
// int は ± 2*10⁹.
// map[]は対数時間．unordered_map[]は一定時間．