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<iomanip>
using namespace std;
int n,m,L;
double p[100];
int t[100],v[100];
double c[51][100];
int main()
{
	cin>>n>>m>>L;
	for(int i=0;i<n;i++)
	{
		cin>>p[i]>>t[i]>>v[i];
		p[i]/=100;
		if(p[i]==1)
		{
			c[m][i]=1;
		}
		else if(p[i]==0)
		{
			c[0][i]=1;
		}
		else
		{
			c[0][i]=1;
			for(int j=0;j<m;j++)c[0][i]*=1-p[i];
			for(int j=0;j<m;j++)c[j+1][i]=c[j][i]*(m-j)/(j+1)/(1/p[i]-1);
		}
	}
	for(int i=0;i<n;i++)
	{
		double ans=0;
		for(int j=0;j<=m;j++)
		{
			double np=c[j][i];
			for(int k=0;k<n;k++)
			{
				if(i==k)continue;
				int cnt=0;
				double P=1;
				while(cnt<=m&&L*v[k]+j*t[i]*v[i]*v[k]>=L*v[i]+cnt*t[k]*v[i]*v[k])
				{
					P-=c[cnt][k];
					cnt++;
				}
				np*=P;
			}
			ans+=np;
		}
		cout<<fixed<<setprecision(9)<<ans<<endl;
	}
}