codeforces285B

Find Marble

CodeForces - 285B

Petya and Vasya are playing a game. Petya's got n non-transparent glasses, standing in a row. The glasses' positions are indexed with integers from 1 to n from left to right. Note that the positions are indexed but the glasses are not.

First Petya puts a marble under the glass in position s. Then he performs some (possibly zero) shuffling operations. One shuffling operation means moving the glass from the first position to position p₁, the glass from the second position to position p₂ and so on. That is, a glass goes from position i to position p_i. Consider all glasses are moving simultaneously during one shuffling operation. When the glasses are shuffled, the marble doesn't travel from one glass to another: it moves together with the glass it was initially been put in.

After all shuffling operations Petya shows Vasya that the ball has moved to position t. Vasya's task is to say what minimum number of shuffling operations Petya has performed or determine that Petya has made a mistake and the marble could not have got from position s to position t.

Input

The first line contains three integers: n, s, t (1 ≤ n ≤ 10⁵; 1 ≤ s, t ≤ n) — the number of glasses, the ball's initial and final position. The second line contains nspace-separated integers: p₁, p₂, ..., p_n (1 ≤ p_i ≤ n) — the shuffling operation parameters. It is guaranteed that all p_i's are distinct.

Note that s can equal t.

Output

If the marble can move from position s to position t, then print on a single line a non-negative integer — the minimum number of shuffling operations, needed to get the marble to position t. If it is impossible, print number -1.

Examples

Input

4 2 1
2 3 4 1

Output

3

Input

4 3 3
4 1 3 2

Output

0

Input

4 3 4
1 2 3 4

Output

-1

Input

3 1 3
2 1 3

Output

-1

sol：容易发现可以交换的最终一定会变成一个环，于是缩成一个个联通块，如果S和T不在同一个块中，就puts("-1")，否则就O(n)模拟需要几步才能到

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=;
    bool f=;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<)+(s<<)+(ch^); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<)
    {
        putchar('-'); x=-x;
    }
    if(x<)
    {
        putchar(x+'');    return;
    }
    write(x/);
    putchar((x%)+'');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int n,S,T,P[N];
int Father[N];
inline int Get_Father(int x)
{
    return (Father[x]==x)?(Father[x]):(Father[x]=Get_Father(Father[x]));
}
inline void Merg(int x,int y)
{
    int xx=Get_Father(x),yy=Get_Father(y);
    if(xx==yy) return;
    Father[xx]=yy;
}
int main()
{
    int i;
    R(n); R(S); R(T);
    for(i=;i<=n;i++) Father[i]=i;
    for(i=;i<=n;i++)
    {
        R(P[i]); Merg(i,P[i]);
    }
    if(Get_Father(S)!=Get_Father(T)) return *puts("-1");
    int Now=S,ans=;
    while(Now!=T)
    {
        Now=P[Now]; ans++;
    }
    Wl(ans);
    return ;
}
/*
input
4 2 1
2 3 4 1
output
3

input
4 3 3
4 1 3 2
output
0

input
4 3 4
1 2 3 4
output
-1

input
3 1 3
2 1 3
output
-1
*/