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;
#define eps (1e-8)

int N,M;
double L;
double P[100],T[100],V[100];
double a[100][100],b[100][100];

double mem[100][100];
double ncr(int n,int r){
  if(mem[n][r]!=-1.0)return mem[n][r];
  r=min(r,n-r);
  double res=1.0;
  for(int i=0;i<r;i++){
    res=res*(double)(n-i);
    res=res/(double)(i+1);
  }
  return mem[n][r]=res;
}

double solve(int x){
  if(V[x]<eps)return 0;
  
  double res=0;
  for(int i=0;i<=M;i++){
    double p=a[x][i];
    double d=b[x][i];
    for(int j=0;j<N;j++){
      if(x==j)continue;
      if(V[j]<eps){
        continue;
      }
      double sum=0;
      for(int k=0;k<=M;k++){
        if(b[j][k]<d+eps)continue;
        sum+=a[j][k];
      }
      p*=sum;
    }
    res+=p;
  }
  return res;
}

int main(){
  for(int i=0;i<100;i++)
    for(int j=0;j<100;j++)
      mem[i][j]=-1.0;
  
  cin>>N>>M>>L;
  for(int i=0;i<N;i++){
    cin>>P[i]>>T[i]>>V[i];
    if(V[i]<eps)continue;

    double p=P[i]/100.0;
    double q=1.0-p;
    for(int j=0;j<=M;j++){
      a[i][j]=ncr(M,j)*pow(p,j)*pow(q,M-j);
      b[i][j]=L/V[i]+T[i]*j;
      //cout<<a[i][j]<<' '<<b[i][j]<<endl;
    }
    //cout<<endl;
  }
  for(int i=0;i<N;i++){
    printf("%.10f\n",solve(i));
  }
  return 0;
}