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)

#define _GLIBCXX_DEBUG
typedef long long ll;
#include <bits/stdc++.h>

using namespace std;

ll mod = 1000000007;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

ll modpow(ll a, ll n, ll mod)
{
	ll res = 1;
	while (0 < n)
	{
		if (n & 1)
		{
			res = res * a % mod;
		}
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

//int gcd(int x, int y)
//{
//
//}

int main()
{
	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
	}

	int before = a[0];
	if (a[0] == 2)
	{
		cout << 1 << endl;
		return 0;
	}
	else
	{
		for (int i = 1; i < n; i++)
		{
			int after = a[before - 1];
			if (after == 2)
			{
				cout << i + 1 << endl;
				return 0;
			}
			before = after;
		}
	}
	cout << -1 << endl;
}