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<deque>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include<algorithm>
#include<string>
#include<iostream>
#include<sstream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
long long cm(int n,int m){  
  m=n-m<m?n-m:m;
  long long a=1;
  int b=n-m+1,c=1;
  for(;c<=m;){
    a=a*b/c;
    ++b;
    ++c;
  }
  return a;
}
int main(){
  int i,j,k;
  int n,m,l;
  scanf("%d%d%d",&n,&m,&l);
  int a[100],b[100],c[100];
  for(i=0;i<n;++i)
    scanf("%d%d%d",a+i,b+i,c+i);
  long long d[51];
  for(i=0;i<=(m+1)/2;++i)
    d[i]=d[m-i]=cm(m,i);
  double e[100][51],f[100][51],g[100][52];
  for(i=0;i<n;++i){
    g[i][m+1]=0;
    for(j=m;j>=0;--j){
      e[i][j]=(double)l/c[i]+b[i]*j;
      f[i][j]=pow((double)a[i]/100,j)*pow((double)(100-a[i])/100,m-j)*d[j];
      g[i][j]=g[i][j+1]+f[i][j];
    }
  }
  for(i=0;i<n;++i){
    double pr=0;
    for(j=0;j<=m;++j){
      double pi=f[i][j];
      for(k=0;k<n;++k){
	if(i!=k)
	  pi*=g[k][upper_bound(e[k],e[k]+m+1,e[i][j])-e[k]];
      }
      pr+=pi;
    }
    printf("%.8f\n",pr);
  }
  return 0;
}