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<stdio.h>
#include<cstring>
#include<iostream>
#include<string>

using namespace std;

const double eps = 1e-12;
#define REP(i, x) for (int i=0;i<x;i++)

double tmp[107][107];
double rest[107][107],rsum[107][107];
int N,M,L;
int p[107],t[107],v[107];
double pp[107];

int main()
{
    while (scanf("%d%d%d",&N,&M,&L) != EOF)
    {
        REP(i,N) scanf("%d%d%d",&p[i],&t[i],&v[i]);
        REP(i,N) pp[i] = p[i] / 100.0;
        REP(i,N)
        {
            REP(j,M+1) REP(k,M+1) tmp[j][k] = 0;
            tmp[0][0] = 1.00;
            for (int j = 1;j <= M;++j)
                for (int k = 0;k <= j;++k)
                {
                    tmp[j][k] = tmp[j-1][k] * (1-pp[i]);
                    if (k) tmp[j][k] += tmp[j-1][k-1] * pp[i];
                }
            for (int j = M;j >= 0;--j)
            {
                rest[i][j] = tmp[M][j];
                rsum[i][j] = tmp[M][j];
           //     cout<<rest[i][j]<<" ";
                if (j!=M) rsum[i][j] += rsum[i][j+1];
            }
        }
        REP(i,N)
        {
            double ans = 0;
            for (int times = 0;times <= M && v[i] != 0;++times)
            {
                double part = 1.00;
                double use = L / (double)v[i] + t[i] * times;
                for (int k = 0;k < N;++k)
                    if (k != i)
                    {
                        if (v[k] == 0) continue;
                        int l = 0,r = M;
                        while (l < r)
                        {
                            int mid = (l+r)>>1;
                            double total = L / (double) v[k] + t[k] * mid;
                            if (total <= use)
                                l = mid+1;
                            else r = mid;
                        }
                        double total = L / (double) v[k] + t[k] * l;
                        
                      //  --l;
                        if (total <= use)
                        {
                            part = 0.00;
                            break;
                        } else
                            part *= rsum[k][l];
                    }
                ans += rest[i][times] * part;
            }
            printf("%.8f\n",ans);
        }
    }
    return 0;
}