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 *p*<sub class="lower-index">1</sub>, the glass from the second position to position *p*<sub class="lower-index">2</sub> and so on. That is, a glass goes from position *i* to position *p*<sub class="lower-index">*i*</sub>. 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*.
The first line contains three integers: *n*, *s*, *t* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>; 1 ≤ *s*, *t* ≤ *n*) — the number of glasses, the ball's initial and final position. The second line contains *n* space-separated integers: *p*<sub class="lower-index">1</sub>, *p*<sub class="lower-index">2</sub>, ..., *p*<sub class="lower-index">*n*</sub> (1 ≤ *p*<sub class="lower-index">*i*</sub> ≤ *n*) — the shuffling operation parameters. It is guaranteed that all *p*<sub class="lower-index">*i*</sub>'s are distinct.

Note that *s* can equal *t*.
# -*- coding: utf-8 -*-
"""Untitled65.ipynb

Automatically generated by Colaboratory.

Original file is located at
    https://colab.research.google.com/drive/1kH9gF2Sz0uEW-8MC5Exb7nZuB4f5BTlb
"""

n,s,t=map(int,input().split())
l1=list(map(int,input().split()))
l2=[]
l3=[]
for k in range(1,n+1):
  l2.append(k)
  l3.append(0)
c=1
i=0
if s==t:
  print("0")
else:
  while l1[s-1]!=l3[t-1] and i<n:
    c=c+1
    for j in range(0,n):
      m=l1[j]-1
      l3[j]=l2[m]
    l2=[]
    for x in l3:
      l2.append(x)
    i=i+1
  if i==n:
    print("-1")
  else:
    print(c)

