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 <vector>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int inf = 0xfffffff;


double p[101][55],T[101][55];
int n,m;
double l;
double c[55][55],pp[101],t[101],v[101];
void init()
{
	c[0][0]=1;
	for(int i=1;i<55;i++){
		c[i][0]=1;
		for(int j=1;j<=i;j++){
			c[i][j]=c[i-1][j]+c[i-1][j-1];
		}
	}
	for(int i=0;i<=m;i++){
		for(int j=0;j<n;j++){
			p[j][i]=c[m][i]*pow(pp[j],i)*pow((1-pp[j]),m-i);
			T[j][i]=l/v[j]+i*t[j];
			//cout<<j<<" "<<i<<" "<<p[j][i]<<" "<<T[j][i]<<endl;
		}
	}
}
double Ans[101];
double tmp=0;
void dfs(int cur,double cmp,int N,double P){
	if(N==n){
		tmp=P;
		return ;
	}
	if(cur==N){
		dfs(cur,cmp,N+1,P);return ;
	}
	double O=0;
	for(int i=0;i<=m;i++){
		if(T[N][i]>cmp)O+=p[N][i];
	}
	//cout<<O<<" "<<N<<endl;
	dfs(cur,cmp,N+1,P*O);
}
int main()
{
	cin>>n>>m>>l;
	for(int i=0;i<n;i++){
		scanf("%lf%lf%lf",&pp[i],&t[i],&v[i]);
		pp[i]/=100.0;
	}
	init();
	for(int i=0;i<n;i++){
		for(int j=0;j<=m;j++){
			tmp=0;
			dfs(i,T[i][j],0,p[i][j]);
			//cout<<tmp<<" "<<T[i][j]<<" "<<p[i][j]<<endl;
			Ans[i]+=tmp;
		}
		printf("%.10f\n",Ans[i]);
	}
}