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 <iostream>
#include <iomanip>
#include <vector>
#include <cmath>

using namespace std;

double choose(int n, int k){
  double res = 1.;
  for(int i=0;i<k;++i)
	res *= (n - i) * 1. / (k - i);
  return res;
}

double bin(double p, int M, int x, int T){
  double res = 0.;
  for(int i=0;i<=M;++i)
	res += choose(M,i) * pow(p,i) * pow(1-p,M-i) * i * T;
  return res;
}

int main(){
  int N, M, L; cin >> N >> M >> L;
  vector<int> P(N), T(N), V(N);
  for(int i=0;i<N;++i){
	cin >> P[i] >> T[i] >> V[i];
  }

  double S[110][100], prob[110][100];
  vector<double> ans(N,0);
  for(int i=0;i<N;++i){
	for(int j=0;j<=M;++j){
	  prob[i][j] = choose(M,j) * pow(P[i]/100.,j) * pow((100-P[i])/100.,M-j);
	  S[i][j] = L *1./V[i] + j*T[i];
	}
  }

  for(int i=0;i<N;++i)
	for(int j=0;j<=M;++j){
	  double p = 1.;
	  for(int i1=0;i1<N;++i1){
		if(i == i1) continue;
		double q = 0.;
		for(int j1=0;j1<=M;++j1){
		  if(S[i1][j1] > S[i][j])
			q += prob[i1][j1];
		}
		p *= q;
	  }
	  ans[i] += p * prob[i][j];
	}
  for(int i=0;i<N;++i)
	cout << fixed << setprecision(9) << ans[i] << endl;
  

  return 0;
}