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<cmath>
using namespace std;

int main(){
  int n,m,l;
  double p[100];
  int t[100],v[100];
  double time[100][51],prob[100][51];

  cin >> n >> m >> l;
  for(int i=0;i<n;i++)cin >> p[i] >> t[i] >> v[i];

  
  for(int i=0;i<n;i++){
    for(int j=0;j<=m;j++)time[i][j] = (double)l/v[i] + t[i]*j;
  }
  
  double k = 1.0;
  for(int i=0;i<m;i++)k*=100.0;

  for(int i=0;i<n;i++){
    prob[i][0] = pow((double)(100-p[i]),m);
    prob[i][0] /= k;
    double c = m;
    for(int j=1;j<=m;j++){
      prob[i][j] = c * pow((double)(100-p[i]),m-j) * pow((double)p[i],j);
      prob[i][j] /= k;
      c *= m-j;
      c /= j+1;
    }
  }
  /*
  for(int i=0;i<n;i++){
    for(int j=0;j<=m;j++){
      if(prob[i][j] < 0)cout << i << " " << j << endl;;
    }
  }
  */
  for(int i=0;i<n;i++){
    double ans = 0.0;
    if(!v[i])printf("%.8lf\n",ans);
    else{
      for(int j=0;j<=m;j++){
	double tmp = 1.0;
	for(int k=0;k<n;k++){
	  if(k!=i){
	    double hoge = 0.0;
	    for(int l=m;l>=0;l--){
	      if(time[k][l] > time[i][j]){
		hoge += prob[k][l];
	      }
	    }
	    tmp *= hoge;
	  }
	}
	ans += tmp*prob[i][j];
      }
      printf("%.8lf\n",ans);
    }
  }
}