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 <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <cfloat>

using std::cerr;
using std::cin;
using std::cout;
using std::endl;

struct Runner{
  int p,t,v;
  Runner(int _p,int _t,int _v):p(_p),v(_v),t(_t){

  }
  Runner(){}
};

double RepeatedPow(double base, int64_t exponent) {
  if (exponent == 0)
    return 1;
  else if (exponent % 2 == 0) {
    double root = RepeatedPow(base, exponent / 2);
    return (root * root);
  } else {
    return (base * RepeatedPow(base, exponent - 1));
  }
}

int64_t ncr[101][101]={};

double probability(int m,int p, int a){
  double result=1;
  result*=ncr[m][a];
  result*=pow(p/(double)100,a);
  result*=pow((1-p/(double)100),m-a);
  return result;
}

int main(){
  cout << std::fixed << std::setprecision(16);
  cin.tie(nullptr);
  std::ios::sync_with_stdio(false);

  int n,m,l;
  cin>>n>>m>>l;



  for(int i=0;i<=100;i++){
    for(int j=0;j<=i;j++){
      if(j==0||j==i){
        ncr[i][j]=1;
      }else{
        ncr[i][j]=ncr[i-1][j-1]+ncr[i-1][j];
      }
    }
  }

  std::vector<Runner> runner(n);
  for(int i=0;i<n;i++){
    int p, t, v;
    cin >> p >> t >> v;
    runner[i]=Runner(p,t,v);
  }

  for(int i=0;i<n;i++){
    double result=0;
    for(int a=0;a<=m;a++){
      double term=probability(m,runner[i].p,a);
      double i_time=(double)l/runner[i].v+a*runner[i].t;
      for(int b=0;b<n;b++){
        if(i==b) continue;
        double win_rate=0;
        for(int c=m;c>=0;c--){
          double probability_b=probability(m,runner[b].p,c);
          double b_time=(double)l/runner[b].v+c*runner[b].t;
          if(i_time<b_time){
            win_rate+=probability_b;
          }else{
            break;
          }
        }
        term*=win_rate;
      }
      result+=term;
    }
    cout<<result<<endl;
  }

  return 0;
}