About Cyclic Permutation

I studied math, and I come up with this problem. There are two permutations A and B and a integer M. We say A almost equal to B if we can make from A to B doing following operations.
-1 Choose a M-length segment of the permutation A.
-2 Perform a cyclic shift on it towards the right.(So,if sub segment is "1 2 3 4 5"(m=5), then after this operation , it will be "5 1 2 3 4".)

Question : Does A almost equal to B?

I think it is typical , but I couldn't find the answer. How to solve it?(not brute force!)

number of elements in the permutation<=10^5

example

A "1 2 3"
B "2 3 1"
m=2
answer　YES
explain　"1 2 3"->"2 1 3"->"2 3 1"

Here's a proof of my conjecture. Let n be the length of the permutations and m be the length of the windows that we are allowed to rotate, where 1 ≤ m ≤ n. Permutations P and Q are almost equal if there exists a sequence of window rotations that transforms P into Q. Almost equality is an equivalence relation. Here's the claimed characterization of the equivalence classes.

(1) m = 1: P almost equals Q if and only if P = Q
(2) m = n: P almost equals Q if and only if they're rotations of each other
(3) 1 < m < n, m odd: P almost equals Q if and only if they have the same parity
(4) 1 < m < n, n even: P almost equals Q

The first two claims are obvious. As for (3), the necessity of the parity condition follows from the fact that rotating a window of odd length is an even permutation.

The meat of the argument here is to find an algorithm for when n = m + 1 ≥ 4, since in general, we can use an algorithm similar to insertion sort to transform P so that all but the last m + 1 elements match Q, and specifically, the case (n, m) = (3, 2) can be solved by inspection. In case m is even, we ensure further that the transformation matches the parity of Q, by rotating the last m elements once if necessary. (For m odd, we just assume equal parities.)

We need a technique for moving fewer than m elements at once. Suppose that the state is as follows.

1, 2, 3, 4, ..., m, m + 1

Rotate the second window m - 1 times (i.e., once in reverse).

1, 3, 4, ..., m, m + 1, 2

Rotate the first window m - 1 times.

3, 4, ..., m, m + 1, 1, 2

Second, twice.

3, 2, 4, ..., m, m + 1, 1
3, 1, 2, 4, ..., m, m + 1

We've succeeded in rotating the first three elements. This suffices in combination with conjugation by rotations to "insertion sort" the first m - 1 elements of Q into place. The other two are in the right order by the parity match.