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 <algorithm>
#include <string>
#include <queue>
#include <cstdio>
#include <cstdlib>
#include <complex>
#include <cstring>
#include <cmath>
#include <string>
using namespace std;
#define min(a,b) a<b?a:b;
#define max(a,b) a>b?a:b;
#define LL long long
const int N = 210;
const double eps = 1e-8;
double dp[N][N],p[N][N],ans[N];//莠コ?御シ第?蜃?惻縲よ慮髣エ?梧ヲら紫
double zu[N][N];
int n;
struct node{
	double p,t,v;
}data[N];
void zuhe()
{
	memset(zu,0,sizeof(zu));
	int i,j;
	for(i=0;i<=50;i++)
	{
		zu[i][0]=1;
		zu[i][1]=i;
	}
	for(i=2;i<=50;i++)
		for(j=1;j<=i;j++)
			zu[i][j]=zu[i-1][j]+zu[i-1][j-1];
}
int main()
{
	int i,j,z,k,m,l;
	double sum,res,time1,time2;
	zuhe();
	while(scanf("%d%d%d",&n,&m,&l)!=EOF)
	{
		for(i=1;i<=n;i++)
			scanf("%lf%lf%lf",&data[i].p,&data[i].t,&data[i].v);
		for(i=1;i<=n;i++)
			for(j=0;j<=m;j++)
				p[i][j]=zu[m][j]*pow(data[i].p/100.0,j*1.0)*pow((1-data[i].p/100.0),(m-j)*1.0);
		for(i=1;i<=n;i++)
		{
			ans[i]=0;
			for(j=0;j<=m;j++)
			{
				res=p[i][j];
				time1=j*data[i].t+l/data[i].v;
				for(z=1;z<=n;z++)
				{
					sum=0;
					if(z==i)
						continue;
					for(k=0;k<=m;k++)
					{
						time2=k*data[z].t+l/data[z].v;
						if(time1<time2-eps)
							sum+=p[z][k];
					}
					res*=sum;
				}
				ans[i]+=res;
			}
		}
		for(i=1;i<=n;i++)
			printf("%.8lf\n",ans[i]);
	}
	return 0;
}



/*
2 2 50
30 50 1
30 50 2

2 1 100
100 100 10
0 100 1

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

2 2 50
30 0 1
30 50 2

  */