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 r(i,n) for(int i=0;i<n;i++)
using namespace std;

int n,m;
double l,p[111],t[111],v[111],ans;
double dp[111][111][111],T[111][111];

int main(){
  cin>>n>>m>>l;
  r(i,n)cin>>p[i]>>t[i]>>v[i],p[i]/=100;
  r(i,n){
    dp[i][0][0]=1;
    r(j,m+1){
      if(j!=m)r(k,j+1){
        dp[i][j+1][k+1]+=dp[i][j][k]*p[i];
        dp[i][j+1][k]+=dp[i][j][k]*(1.0-p[i]);
      }
      if(v[i]>0)T[i][j]=l/v[i]+t[i]*j;
      else T[i][j]=1e9;
    }
  }
  r(i,n){
    ans=0;
    if(v[i]==0){
      cout<<0<<endl;
      continue;
    }
    r(j,m+1){
      double p=dp[i][m][j];
      r(k,n)if(k!=i){
        double sum=0;
        r(x,m+1) if(T[i][j]<T[k][x]) sum+=dp[k][m][x];
        p*=sum;
      }
      ans+=p;
    }
    printf("%.9f\n",ans);
  }
}