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<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<ctime>
#include<vector>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long LL;
const double pi=3.141592653;
const double eps=1e-10;
const int INF=0x3f3f3f3f;
const int MOD=10007;
const int MAXN=220;
struct people
{
    double pp[MAXN],time[MAXN];
} peo[MAXN];

int n,m,L,p[MAXN],t[MAXN],v[MAXN];
LL Com[MAXN][MAXN];

int main()
{
    Com[0][0]=1;
    for(int i=1;i<=50;i++)
    {
        Com[i][0]=Com[i-1][0];
        for(int j=1;j<=i;j++)
            Com[i][j]=Com[i-1][j]+Com[i-1][j-1];
    }
    while(scanf("%d%d%d",&n,&m,&L)!=EOF)
    {
        for(int i=0; i<n; i++)
            scanf("%d%d%d",&p[i],&t[i],&v[i]);
        for(int i=0; i<n; i++)
            for(int j=0; j<=m; j++)
            {
                double x=Com[m][j];
                int k;
                for(k=1;k<=j;k++)
                    x*=p[i]/100.0;
                for(;k<=m;k++)
                    x*=(100-p[i])/100.0;
                peo[i].pp[j]=x;
                peo[i].time[j]=(L+0.0)/v[i]+j*t[i];
            }
        for(int i=0; i<n; i++)
        {
            double ans=0;
            for(int j=0; j<=m; j++)
            {
                double tmp=1;
                for(int k=0; k<n; k++)
                {
                    if(k==i)
                        continue;
                    double x=0;
                    for(int l=0; l<=m; l++)
                        if(peo[k].time[l]>peo[i].time[j])
                            x+=peo[k].pp[l];
                    tmp*=x;
                }
                ans+=tmp*peo[i].pp[j];
            }
            printf("%.8lf\n",ans);
        }
    }
    return 0;
}