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<bits/stdc++.h>
using namespace std;
double C(int x,int y){
  double res=1;
  for(int i=0;i<y;i++) res*=x-i;
  for(int i=1;i<=y;i++) res/=i;
  return res;
}
int main(){
  int n,m;
  double l;
  cin >> n >> m >> l;
  int i,j,k,a;
  double p[n],t[n],v[n];
  double b[n][m+1];
  for(i=0;i<n;i++) {
    cin >> p[i] >> t[i] >> v[i];
    for(j=0;j<=m;j++){
      b[i][j]=C(m,j)*pow(p[i]/100,j)*pow(1.0-p[i]/100,m-j);
    }
  }
  double ans,x,y;
  for(i=0;i<n;i++){
    ans=0;
    for(j=0;j<=m;j++){
      x=b[i][j];
      for(k=0;k<n;k++){
    if(i==k) continue;
    y=0;
    for(a=m;a>=0;a--){
      if(l/v[k]+t[k]*a>l/v[i]+t[i]*j) y+=b[k][a];
      else break;
    }
    x*=y;
      }
      ans+=x;
    }
    printf("%.8f\n",ans);
  }
  return 0;
}