Codeforces Round#175 (Div.2) 275B

B. Find Marble

time limit per test	2 seconds
memory limit per test	256 megabytes
input	standard input
output	standard output

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 n space-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.

Sample test(s)

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

解题思路：
这道题就是一个过程模拟的题，虽然题目中提到每次执行的时候所有的玻璃杯都会移动，但是我们可以看到的是滚珠(marble)所在的杯子是
不会更换的，因此我们可以讲此过程简化为marble的位置变化模拟。由于玻璃杯上的序号是不变的 并且互不相同，因此玻璃杯若能找到t，则
模拟的过程中不能出现环，也就是说所经过的次数不会超过玻璃杯的总数n，故代码如下：