Sheriff gave all the bandits numbers from 1 to N. For each place i he determined the unique position j. After whistling the bandit staying on position i should run to the j-th position. Sheriff loved seeing how the bandits move around, and he continued whistling until the evening. He finished the game only when he noticed that the bandits are in the same order in which they were standing originally.

Now the Sheriff asks the question: How many times has he whistled?

Input:

- The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains a single integer N denoting the number of bandits. The second line contains N space-separated integers A1, A2, …, AN denoting that the bandit staying on position i should run to the Ai-th position after the whistle.

2

3

1 2 3

5

2 3 1 5 4

Output:

1

6

Logic :

Method 1) You can create a function where you pass initial arrangement ie 2 3 1 5 4.

in loop you can create result array first as 1 2 3 4 5 and generate new arrangement array ie 3 1 2 5 4 and check whether its matching with resultant array 1 2 3 4 5 or not , again with same indexing create next arrangement ie 2 3 1 4 5 and check … and so on, Keep one counter when both matches return count.

Method 2) You can optimize space with just using one array Initial and search whether all are in increasing order ( 1,2,3,4,5) then return count.

Pass array of initial arrangement ie 2 3 1 5 4 and Integers ( 1 to 5 ) to a function sgarden. Then calculate that each integer is getting its own value After how many iteration of rearrangement.

Such way calculate for each integer (1 to 5) and find lcm of this count which will be total number of rearrangement needed.

example : 1,2,3 are getting its own place after 3 iteration. While 4 and 5 after 2.

So whole array will get rearranged after 6 iterations.

**Full coverage of rearrangements**

the bandits positions are:

0. 1 2 3 4 5

1. 3 1 2 5 4

2. 2 3 1 4 5

3. 1 2 3 5 4

4. 3 1 2 4 5

5. 2 3 1 5 4

6. 1 2 3 4 5.

PS : for understanding and for loop’s indexing purpose we have decremented initial array 2 3 1 5 4 by 1. and hence passing 1 2 0 4 3.

Thanks to Dhaval for suggesting this approach and Article

Code

__You can see Working code at : http://ideone.com/qXGU6w__