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 <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
const double eps=1e-7;
const int maxn=110;
const int maxm=110;
double lin[maxm][maxm],tmp[maxm][maxm],c[maxm][maxm],ans[maxm],p[maxm],v[maxm],t[maxm];
int main()
{
	for(int i=0;i<maxm;i++)
	{
		c[i][0]=c[i][i]=1;
		for(int j=1;j<i;j++) c[i][j]=c[i-1][j]+c[i-1][j-1];
	}
	int n,m;
	double l;
	while(scanf("%d%d%lf",&n,&m,&l)!=EOF)
	{
		for(int i=0;i<n;i++) scanf("%lf%lf%lf",&p[i],&t[i],&v[i]);
		for(int i=0;i<n;i++) p[i]/=100;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<=m;j++)
			{
				tmp[i][j]=l/v[i]+j*t[i];
				lin[i][j]=pow(p[i],j)*pow((1-p[i]),m-j)*c[m][j];
			}
			for(int j=m-1;j>=0;j--) lin[i][j]+=lin[i][j+1];
			//for(int j=0;j<=m;j++) printf("%f ",lin[i][j]);
			//printf("\n");
			lin[i][m+1]=0;	
		}
		for(int i=0;i<n;i++)
		{
			for(int k=0;k<=m;k++) ans[k]=pow(p[i],k)*pow((1-p[i]),m-k)*c[m][k];
			for(int j=0;j<n;j++) 
				if(j!=i)
				{
					int q=m;
					for(int p=m;p>=0;p--)
					{
						while(q>=0 && tmp[i][q]>=tmp[j][p]-eps) 
						{
							ans[q]*=lin[j][p+1];
							q--;
						}
						if(q==-1) break;
					}
				}
			double sum=0;
			for(int k=0;k<=m;k++) sum+=ans[k];
			printf("%.9f\n",sum);
		}
	}
}