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 <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
long long c[55][55];
double dp1[105][52],dp2[105][52];
int main()
{
    for(int i=0;i<=50;++i)
    {
        c[i][0]=c[i][i]=1;
        for(int j=1;j<i;++j) c[i][j]=(c[i-1][j]+c[i-1][j-1]);
    }
    int n,m,L,P[105],T[105],V[105];
    scanf("%d%d%d",&n,&m,&L);
    for(int i=0;i<n;++i) scanf("%d%d%d",&P[i],&T[i],&V[i]);
    for(int i=0;i<n;++i)
    {
        double p[2][55];
        p[0][0]=1.0;
        p[1][0]=1.0;
        for(int j=1;j<=m;++j)
        {
            p[0][j]=p[0][j-1]*P[i]/100;
            p[1][j]=p[1][j-1]*(100-P[i])/100;
        }
        for(int j=0;j<=m;++j)
        {
            dp1[i][j]=c[m][j]*p[0][j]*p[1][m-j];
            dp2[i][j]=L*1.0/V[i]+j*T[i];
        }
    }
    for(int i=0;i<n;++i)
    {
        double ans=0;
        for(int j=0;j<=m;++j)
        {
            double tp=dp1[i][j];
            double lim=dp2[i][j];
            for(int x=0;x<n;++x)
            {
                if(x==i) continue;
                double q=0;
                for(int y=m;y>=0;--y) if(dp2[x][y]>lim) q+=dp1[x][y];
                tp*=q;
            }
            ans+=tp;
        }
        printf("%.8lf\n",ans);
    }
    return 0;
}