Slave to achievements 1

Gyeonggeun plays a web game called isRPG. The title has RPG in it, so as you would expect, the game holds many pieces of equipment a character can use. Every piece is built by gathering materials in a fixed recipe, and the options on a finished piece are drawn at random inside a fixed range. If a finished piece is disappointing, you can take it apart and get some of the spent materials back.

Like most games these days, isRPG has an achievement system: reach a threshold on some counter and the achievement unlocks. The two counters Gyeonggeun is after right now are "craft at least this many pieces of equipment" and "disassemble at least this many pieces of equipment".

The easiest thing to craft in isRPG is the wooden dagger. Its only material is wood chips, and wood chips are easy to buy at the shop. Achievements have nothing to do with the quality of the item, so Gyeonggeun plans to whittle a great many wooden daggers.

Spending $N$ wood chips crafts one wooden dagger. Disassembling one wooden dagger returns at least $0$ and at most $K$ wood chips. The probability of getting $x$ ($0 \le x \le K$) wood chips back is $1/(K+1)$ for every $x$, whatever the value of $x$, and separate disassemblies are independent.

Gyeonggeun killed plenty of treasure-hoarding dragons, and with the loot he bought $M$ wood chips at the shop. He owns no wooden dagger right now. From here on he repeats the following work. First he crafts wooden daggers until he cannot craft another one, then he disassembles every wooden dagger he holds. When he cannot craft even one dagger, the work stops.

Long experience taught Gyeonggeun that wood chips left in the inventory after all the work are very unpleasant. So he wants the probability that $0$ wood chips are left once the work ends. Having thought that far, he also wants the probability that $i$ ($0 \le i < N$) wood chips are left. Help him out.

The first line contains the number of wood chips needed for one wooden dagger $N$, the largest number of wood chips one disassembly returns $K$, and the number of wood chips Gyeonggeun holds right now $M$, separated by spaces. ($1 \le K < N \le 10^3$, $0 \le M \le 10^6$)

Print $N$ lines. Line $i$ holds the probability that $i-1$ wood chips are left. At most $N-1$ wood chips can be left, so the output is exactly $N$ lines.

Write each probability as one integer, not as a decimal. Writing the probability as a reduced fraction $\frac{p}{q}$, print the integer $r$ with $0 \le r < 10^9+7$ that satisfies $p - qr \equiv 0 \pmod{10^9+7}$. Such an $r$ exists and is unique in that range, and this has been proved.

For $N = 4$, $K = 2$, $M = 4$ the answers are $\frac{1}{3}$, $\frac{1}{3}$, $\frac{1}{3}$, $0$ in that order. Four wood chips craft one dagger and leave none, and disassembling that dagger returns $0$, $1$, or $2$ chips, each with probability $\frac{1}{3}$. In every case no further dagger can be crafted, so the work ends there.