Description:

MZL has n cute boys.They are playing a game♂.The game will run in turn

First,System choose an alive player x randomly.Player x will be out of the game.

Then player x will attack all alive players in the game

When a player is attacked,1−p is the probability of he still lives,p is the probability of he dies

Now mzl wants to know：the probability of one player be out of the game and be attacked k times

You need to print the probability mod 258280327 for every k from 0 to n-1

According to Fermat Theory,(x/y) mod 258280327=x*(y^{}^{}^{258280325}) mod 258280327

p will be given in a special way

Input:

The first line of the input contains a single number T, the number of test cases.

Next T lines, each line contains three integer n,x,y.p=x/y

T≤5, n≤2∗10^{3}^{}^{} ,0≤x≤10^{9}^{}^{}^{ },x+1≤y≤10^{9}.

It is guaranteed that y and 258280327 are coprime.

Output:

T lines, every line n numbers: the ans from 0 to n-1

Sample Input:

2 3 33 100 9 23 233

Sample Output:

172186885 210128265 223268793 229582513 70878931 75916746 175250440 21435537 57513225 236405985 111165243 115953819

Hint:

for case 1: The probability of you live and not be attacked is 1/3 The probability of you live and be attacked for one time is: (2/3)*(0.33*0.67+0.67*0.67*(1/2))=8911/30000

Source:

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