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>
#define T 6005
#define N 105
#define M 55
using namespace std;
int n,m,l,p[N],t[N],v[N];
double dp[N][2][T],sum[N][T];
 
int main(){
  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++){
    dp[i][0][0]=1;
    for(int j=0;j<m;j++){
      for(int k=0;k<T;k++){
	if(!dp[i][j%2][k])continue;
	dp[i][(j+1)%2][k]+=dp[i][j%2][k]*(100.0-p[i])/100.0;
	dp[i][(j+1)%2][k+t[i]]+=dp[i][j%2][k]*p[i]/100.0;
      }
      for(int k=0;k<T;k++)dp[i][j%2][k]=0;
    }
  }
  for(int i=0;i<n;i++)
    for(int j=T-2;j>=0;j--)
      sum[i][j]=sum[i][j+1]+dp[i][m%2][j];
  for(int i=0;i<n;i++){
    double ans=0;
    if(v[i])
      for(int j=0;j<T;j++){
	if(!dp[i][m%2][j])continue;
	double x=dp[i][m%2][j];
	double s=j+1.0*l/v[i];
	for(int k=0;k<n;k++){
	  if(i==k||!v[k])continue;
	  int idx=(int)(s-1.0*l/v[k]);
	  if(s-1.0*l/v[k]<0)idx=0;
	  else idx++;
	  if(T<=idx){
	    x=0;
	    break;
	  }
	  x*=sum[k][idx];
	}
	ans+=x;
      }
    printf("%.8f\n",ans);
  }
  return 0;
}