Problem: N people run a marathon. There are M resting places on the way. For each resting place, the i-th runner takes a break with probability P_i percent. When the i-th runner takes a break, he gets rest for T_i time.

The i-th runner runs at constant speed V_i, and the distance of the marathon is L.

You are requested to compute the probability for each runner to win the first place. If a runner arrives at the goal with another person at the same time, they are not considered to win the first place.



Input

A dataset is given in the following format:

N M L
P_1 T_1 V_1
P_2 T_2 V_2
...
P_N T_N V_N


The first line of a dataset contains three integers N (1 \leq N \leq 100), M (0 \leq M \leq 50) and L (1 \leq L \leq 100,000). N is the number of runners. M is the number of resting places. L is the distance of the marathon.

Each of the following N lines contains three integers P_i (0 \leq P_i \leq 100), T_i (0 \leq T_i \leq 100) and V_i (0 \leq V_i \leq 100) describing the i-th runner. P_i is the probability to take a break. T_i is the time of resting. V_i is the speed.

Output

For each runner, you should answer the probability of winning. The i-th line in the output should be the probability that the i-th runner wins the marathon. Each number in the output should not contain an error greater than 10^{-5}.

Examples

Input

2 2 50
30 50 1
30 50 2


Output

0.28770000
0.71230000


Input

2 1 100
100 100 10
0 100 1


Output

0.00000000
1.00000000


Input

3 1 100
50 1 1
50 1 1
50 1 1


Output

0.12500000
0.12500000
0.12500000


Input

2 2 50
30 0 1
30 50 2


Output

0.51000000
0.49000000

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<utility>
#include<cmath>
#include<algorithm>

using namespace std;

long double fact(int x){
	long double r=1;
	while(x){
		r*=x--;
	}
	return r;
}

int main(){
	int n,m,l;
	cin>>n>>m>>l;
	double ap[100][51],acp[100][52]={{}};
	double at[100][51];
	for(int i=0;i<n;i++){
		double p,t,v;
		cin>>p>>t>>v;
		for(int j=0;j<=m;j++){
			at[i][j]=l/v+t*j;// time for athlete i to rest for j times
			ap[i][j]=pow(p/100,j)*pow(1-p/100,m-j)*fact(m)/fact(j)/fact(m-j);// probability to rest for j times
			acp[i][j+1]=acp[i][j]+ap[i][j];// sum of probability of less than j rests
		}
	}
	for(int i=0;i<n;i++){
		double a=0;
		for(int j=0;j<=m;j++){
			double c=ap[i][j];
			for(int k=0;k<n;k++){
				if(i!=k){
					int x=upper_bound(at[k],at[k]+m+1,at[i][j]+1e-9)-at[k];
					c*=1-acp[k][x];
				}
			}
			a+=c;
		}
		cout<<fixed<<a<<endl;   
	}
	return 0;
}