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>
using namespace std;
int p[110];
int t[110];
int v[110];
double dp[110][110];
double prob[110][110];
int main(){
	int a,b,c;
	scanf("%d%d%d",&a,&b,&c);
	for(int i=0;i<a;i++){
		scanf("%d%d%d",p+i,t+i,v+i);
		for(int j=0;j<=b;j++)for(int k=0;k<=b;k++)dp[j][k]=0;
		dp[0][0]=1;
		for(int j=0;j<b;j++){
			for(int k=0;k<=j;k++){
				dp[j+1][k+1]+=dp[j][k]*p[i]/100;
				dp[j+1][k]+=dp[j][k]*(100-p[i])/100;
			}
		}
		for(int j=0;j<=b;j++)prob[i][j]=dp[b][j];
	}
	double EPS=1e-12;
	for(int i=0;i<a;i++){
		double ret=0;
		for(int j=0;j<=b;j++){
			if(!v[i])continue;
			double time=(double)c/v[i]+j*t[i];
			double ks=prob[i][j];
			for(int k=0;k<a;k++){
				if(i==k)continue;
				double tmp=0;
				for(int l=0;l<=b;l++){
					if(v[k]&&time+EPS>(double)c/v[k]+l*t[k])continue;
					tmp+=prob[k][l];
				}
				ks*=tmp;
			}
			ret+=ks;
		}
		printf("%.12f\n",ret);
	}
}