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 <vector>
#include <algorithm>
#include <iomanip>
#include <cmath>
using namespace std;
const double INF = 1e10;
const double EPS = 1e-10;

int main(){
    int n,m;
    double l;
    cin >> n >> m >> l;
    vector<double> p(n), t(n), v(n);
    for(int i=0; i<n; i++){
        cin >> p[i] >> t[i] >> v[i];
        p[i] /= 100.0;
    }

    vector<vector<double> > c(m+1, vector<double>(m+1, 0));
    for(int i=0; i<=m; i++) c[i][0] = 1;
    for(int i=1; i<=m; i++){
        for(int j=1; j<=i; j++){
            c[i][j] = c[i-1][j-1] +c[i-1][j];
        }
    }

    vector<vector<double> > prob(n, vector<double>(m+1)), time(n, vector<double>(m+1));
    for(int i=0; i<n; i++){
        double ttime = l/v[i];
        for(int j=0; j<=m; j++){
            prob[i][j] = pow(p[i], j) *pow(1-p[i], m-j) *c[m][j];
            time[i][j] = ttime;
            ttime += t[i];
        }
    }
    vector<vector<double> > accumprob = prob;
    for(int i=0; i<n; i++){
        for(int j=m-1; j>=0; j--){
            accumprob[i][j] += accumprob[i][j+1];
        }
    }

    cout << fixed;
    cout << setprecision(10);
    for(int i=0; i<n; i++){
        double subans = 0;
        for(int j=0; j<=m; j++){
            double pr = prob[i][j];
            for(int k=0; k<n; k++){
                if(i==k) continue;
                double subpr = 0;
                for(int s=0; s<=m; s++){
                    if(time[k][s] > time[i][j] +EPS){
                        subpr = accumprob[k][s];
                        break;
                    }
                }
                pr *= subpr;
            }
            subans += pr;
        }
        cout << subans << endl;
    }
    return 0;
}