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>
    using namespace std;
    double c[101][101],p[101][51],notp[101][51];
    double t[101],v[101],ans[101];
    int main()
    {
        int n,m,i,j,k,ii;
        double l,sum;
        scanf("%d%d%lf",&n,&m,&l);
        c[0][0]=1;
        for(i=1;i<=100;++i)
        {
            c[i][0]=1;
            for(j=1;j<=i;++j)
                c[i][j]=c[i-1][j-1]+c[i-1][j];
        }
        //cout <<c[100][50]<<endl;
        for(i=0;i<n;++i)
        {
            scanf("%lf%lf%lf",&p[i][1],&t[i],&v[i]);
            p[i][1]/=100;
            notp[i][0]=p[i][0]=1;
            notp[i][1]=1-p[i][1];
            for(j=2;j<=m;++j)
                p[i][j]=p[i][j-1]*p[i][1],notp[i][j]=notp[i][j-1]*notp[i][1];
        }
        double sp,ap,at,tim;
        for(ii=0;ii<n;++ii)
        {
            ans[ii]=0;
            for(i=0;i<=m;++i)
            {
                sp=c[m][i]*p[ii][i]*notp[ii][m-i];
                tim=i*t[ii]+l/v[ii];
                //cout <<ii<<" "<<i<<" "<<sp<<" "<<tim<<endl;
                for(j=0;j<n;++j)
                {
                    if(j==ii)continue;
                    sum=0;
                    for(k=0;k<=m;++k)
                    {
                        ap=c[m][k]*p[j][k]*notp[j][m-k];
                        at=k*t[j]+l/v[j];
                        if(at>tim)
                        {
                            sum+=ap;
                        }
                    }
                    sp*=sum;
                }
                ans[ii]+=sp;
            }
        }
        //cout<<c[50][20]<<" "<<p[2][20]<<" "<<p[2][30]<<" "<<c[50][20]*p[2][20]*p[2][30]<<endl;
        for(i=0;i<n;++i)
        printf("%.5f\n",ans[i]);
        return 0;
    }