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 <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <cmath>
#include <cstdlib>

using namespace std;
struct node
{
    double p,t,v;
}nd[105];
int n,m,L;
double dp[105][55][55];
int main()
{
    //freopen("D.in", "r", stdin);
    //freopen("D.out3", "w", stdout);
    
    while(~scanf("%d%d%d",&n,&m,&L))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf%lf",&nd[i].p,&nd[i].t,&nd[i].v);
            nd[i].p/=100.0;
        }
        for(int i=0;i<n;i++)
        {
            dp[i][0][0]=1;
            for(int j=1;j<=m;j++)
                for(int k=0;k<=j;k++)
                {
                    if(k>0)
                        dp[i][j][k]=dp[i][j-1][k]*(1-nd[i].p)+dp[i][j-1][k-1]*nd[i].p;
                    else dp[i][j][k]=dp[i][j-1][k]*(1-nd[i].p);
                }
        }
        /*for(int i=0;i<n;i++)
        {
            for(int j=0;j<=m;j++)
            {
                for(int k=0;k<=m;k++)
                    printf("%.6f ",dp[i][j][k]);
                puts("");
            }
            puts("");
        }
         */
        for(int i=0;i<n;i++)
        {
            double ans=0;
            for(int j=0;j<=m;j++)
            {
                double ans1=dp[i][m][j],t=L/(nd[i].v)+j*nd[i].t;
                for(int ii=0;ii<n;ii++)
                {
                    if(ii==i)continue;
                    double tem=0,tt=L/nd[ii].v;
                    for(int jj=0;jj<=m;jj++)
                    {
                        if(tt+jj*(nd[ii].t)>t)
                            break;
                        tem+=dp[ii][m][jj];
                    }
                    ans1*=(1.0-tem);
                }
                ans+=ans1;
            }
            printf("%.8f\n",ans);
        }
    }
    return 0;
}