human
As you have noticed, there are lovely girls in Arpa’s land.

People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.


Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.

The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.

Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.

Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).

Input

The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.

The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.

Output

If there is no t satisfying the condition, print -1. Otherwise print such smallest t.

Examples

Input

4
2 3 1 4

Output

3

Input

4
4 4 4 4

Output

-1

Input

4
2 1 4 3

Output

1

Note

In the first sample suppose t = 3. 

If the first person starts some round:

The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.

The process is similar for the second and the third person.

If the fourth...
Using python can you solve the prior task?
assistant
#Here is Python code to find such t for Mehrdad if it's possible.

#!/usr/bin/python2
# -*- coding: utf-8 -*-

import sys
from fractions import gcd

def rl(proc=None):
    if proc is not None:
        return proc(sys.stdin.readline())
    else:
        return sys.stdin.readline().rstrip()

def srl(proc=None):
    if proc is not None:
        return map(proc, rl().split())
    else:
        return rl().split()

def flow(A, vis, i):
    j = i
    r = 0
    while not vis[j]:
        r += 1
        vis[j] = True
        j = A[j]
    return r if j == i else None

def lcm(numbers):
    def lcm(a, b):
        return (a * b) / gcd(a, b)
    return reduce(lcm, numbers, 1)

def main():
    N = rl(int)
    A = [x-1 for x in srl(int)]
    vis = [False] * len(A)
    B = []
    for i in xrange(len(A)):
        if not vis[i]:
            n = flow(A, vis, i)
            if n is None:
                print -1
                return
            if n % 2 == 0:
                n /= 2
            B.append(n)
    print lcm(B)

if __name__ == '__main__':
    main()