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<cmath>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
int N,M,L;
struct node
{   double t,p;
}per[101][51];
long long c[51][51]={0};
void prepare()
{   c[0][0]=1;c[1][0]=1;
    for(int i=1;i<51;i++)
    for(int j=0;j<51;j++)
    {   if(j==0) c[i][j]=1;
        else c[i][j]=c[i-1][j]+c[i-1][j-1];
    }
}
int main()
{   int T;
    prepare();
    cin>>N>>M>>L;
    for(int i=0;i<N;i++)
    {   double p,t,v;
        cin>>p>>t>>v;p*=0.01;
        per[i][0].t=L/v;
        per[i][0].p=pow(1-p,M);
        for(int j=1;j<=M;j++)
        {   per[i][j].p=c[M][j]*pow(1-p,M-j)*pow(p,j);
            per[i][j].t=per[i][j-1].t+t;
        }
    }
//    for(int i=0;i<N;i++){
//    for(int j=0;j<=M;j++)
//    {   cout<<per[i][j].t<<'('<<per[i][j].p<<')'<<' ';
//
//    }puts("");}
    for(int i=0;i<N;i++)
    {   double s=0;
        for(int j=0;j<=M;j++)
        {   double tt=per[i][j].p;
            for(int k=0;k<N;k++)
            if(k!=i)
            {   double ts=0;
                for(int t=M;t>=0 && per[k][t].t>per[i][j].t;t--)
                {   ts+=per[k][t].p;
                }
                tt*=ts;
               // cout<<ts<<endl;
            }
            s+=tt;
        }
        printf("%.8lf\n",s);
    }
}