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 crush i . 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 crush x and says: " Oww...wwf " (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crush x calls crush crush x 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. crush i = i ). 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 crush i ( 1 ≤ crush i ≤ n ) — the number of i -th person's crush. If there is no t satisfying the condition, print -1 . Otherwise print such smallest t . 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 person starts some round: The fourth person calls hi