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 <vector>
#include <bitset>
#include <map>
#include <queue>
#include <climits>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <cfloat>
using namespace std;

const double INF = DBL_MAX / 2;
const double EPS = 1.0e-10;

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];

	vector<vector<double> > time(n, vector<double>(m+2, INF));
	vector<vector<double> > prob(n, vector<double>(m+1, 0.0)), sumProb(n, vector<double>(m+2, 0.0));
	for(int i=0; i<n; ++i){
		double comb = 1.0;
		for(int j=m; j>=0; --j){
			time[i][j] = l / (double) v[i] + j * t[i];
			prob[i][j] = comb * pow(p[i]/100.0, j) * pow(1.0-p[i]/100.0, m-j);
			sumProb[i][j] = sumProb[i][j+1] + prob[i][j];
			comb *= j;
			comb /= m-j+1;
		}
	}

	for(int i=0; i<n; ++i){
		double ret = 0.0;
		for(int j=0; j<=m; ++j){
			double tmp = prob[i][j];
			for(int k=0; k<n; ++k){
				if(i == k)
					continue;
				int left = 0;
				int right = m+1;
				while(left < right){
					int middle = (left + right) / 2;
					if(time[i][j] < time[k][middle] - EPS){
						right = middle;
					}else{
						left = middle + 1;
					}
				}
				if(time[i][j] < time[k][left] - EPS)
					tmp *= sumProb[k][left];
				else
					tmp *= 0.0;
			}
			ret += tmp;
		}
		printf("%.10f\n", ret);
	}

	return 0;
}