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: Time Limit Exceeded)

#include<bits/stdc++.h>


using namespace std;
#define ll  long long
#define ld  long double
#define endl '\n'
#define FAST ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
const ll mod = 1e9 + 7;
const double PI = 3.141592653589793238463;
const int MAXN = 2e5 + 5;
bool isPrime[MAXN];



ll fastPow(ll x, ll b)
{
    if (b == 0)
        return 1;
    else if (!(b & 1))
    {
        return fastPow(x * x, b / 2);
    }
    else
        return x * fastPow(x * x, (b - 1) / 2);
}

//__builtin_popcount ();
bool isPrim(int n)
{
    if (n == 1 || n == 0)
        return false;
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}



int main()
{
    FAST

    int n;
    cin>>n;
    int arr[n+1];
    for(int i=1;i<=n;i++)
        cin>>arr[i];
    int pt=1;
    int ans=0;
    if(arr[pt]==2)
        return cout<<1,0;
    do{
        pt=arr[pt];
        ans++;
    }while(arr[1]!=arr[pt] && arr[pt]!=2);
    if(pt==1)
        cout<<"-1"<<endl;
    else
        cout<<ans+1<<endl;

    return 0;
}