Rotate and Shift

**Note: The time limit for this problem is 4s, 2x the default.**

To celebrate the start of spring, Farmer John's $N$ cows ($1\le N\le 2\cdot 10^5$) have invented an intriguing new dance, where they stand in a circle and re-order themselves in a predictable way.

Specifically, there are $N$ positions around the circle, numbered sequentially from $0$ to $N-1$, with position $0$ following position $N-1$. A cow resides at each position. The cows are also numbered sequentially from $0$ to $N-1$. Initially, cow $i$ starts in position $i$. You are told a set of $K$ positions $0=A_1<A_2<\dots <A_K<N$ that are "active", meaning the cows in these positions are the next to move ($1\le K\le N$).

In each minute of the dance, two things happen. First, the cows in the active positions rotate: the cow at position $A_1$ moves to position $A_2$, the cow at position $A_2$ moves to position $A_3$, and so on, with the cow at position $A_K$ moving to position $A_1$. All of these $K$ moves happen simultaneously, so the after the rotation is complete, all of the active positions still contain exactly one cow. Next, the active positions themselves shift: $A_1$ becomes $A_1+1$, $A_2$ becomes $A_2+1$, and so on (if $A_i=N-1$ for some active position, then $A_i$ circles back around to $0$).

Please calculate the order of the cows after $T$ minutes of the dance ($1\le T\le 10^9$).

For the example above, here are the cow orders and $A$ for the first four timesteps:

Initial, T = 0: order = [0 1 2 3 4], A = [0 2 3]
T = 1: order = [3 1 0 2 4]
T = 1: A = [1 3 4]
T = 2: order = [3 4 0 1 2]
T = 2: A = [2 4 0]
T = 3: order = [2 4 3 1 0]
T = 3: A = [3 0 1]
T = 4: order = [1 2 3 4 0]

Problem credits: Claire Zhang

Source: USA Computing Olympiad.

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