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<math.h>
#include<algorithm>
using namespace std;
int N,M,L;
double a[105][105],b[105][105],t[105][105],P,T,V;

long double f(int x)
{
	long double r=1;
	for(;x;)r*=x--;
	return r;
}

int main()
{
	scanf("%d%d%d",&N,&M,&L);
	for(int i=0;i<N;i++)
	{
		scanf("%lf%lf%lf",&P,&T,&V);
		for(int j=0;j<=M;j++)
		{
			t[i][j]=L/V+T*j;
			a[i][j]=pow(P/100,j)*pow(1-P/100,M-j)*f(M)/f(j)/f(M-j);
			b[i][j+1]=b[i][j]+a[i][j];
		}
	}
	for(int i=0;i<N;i++)
	{
		double n=0;
		for(int j=0;j<=M;j++)
		{
			double c=a[i][j];
			for(int k=0;k<N;k++)
				if(i!=k)
				{
					int x=upper_bound(t[k],t[k]+M+1,t[i][j]+1e-9)-t[k];
					c*=1-b[k][x];
				}
			n+=c;
		}
		printf("%.8f\n",n);
	}
	return 0;
}