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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int N,M,L;
double X[105];
int T[105],V[105];
double C[105][105];
double P[105][55];
double rest[105][55];
double ans[105];
inline void calC()
{
	int i,j;C[0][0]=1;C[1][0]=1;C[1][1]=1;
	for(i=2;i<=50;i++)
	{
		C[i][i]=1;C[i][0]=1;
		for(j=1;j<i;j++)
			C[i][j]=C[i-1][j]+C[i-1][j-1];
	}
}
inline void calR()
{
	int i,j;
	for(i=1;i<=N;i++)
	{
		rest[i][0]=L*1.0/(V[i]*1.0);
		for(j=1;j<=M;j++)
			rest[i][j]=rest[i][j-1]+T[i];
	}
}
inline void calP()
{
	int i,j;
	for(i=1;i<=N;i++)
		for(j=0;j<=M;j++)
			P[i][j]=C[M][j]*pow(X[i],j*1.0)*pow((1-X[i]),(M-j)*1.0);
}
inline double cal(int pos,double MAXN)
{
	int i,j;double temp;
	double ret=1.0;
	for(i=1;i<=N;i++)
	{
		if(i==pos)
			continue;
		temp=0.0;
		for(j=M;j>=0&&rest[i][j]>MAXN;j--)
			temp+=P[i][j];
		ret*=temp;
	}
	return ret;
}
inline void solve()
{
	int i,j;
	calC();calR();calP();
	for(i=1;i<=N;i++)
	{
		ans[i]=0.0;double temp=1.0;
		for(j=0;j<=M;j++)
			ans[i]+=P[i][j]*cal(i,rest[i][j]);
	}
	for(i=1;i<=N;i++)
		printf("%.7lf\n",ans[i]);
}
int main()
{
	int i,a,b,c;
	scanf("%d%d%d",&N,&M,&L);	
	for(i=1;i<=N;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		X[i]=a*1.0/100;T[i]=b;V[i]=c;
	}
	solve();
	return 0;
}