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>
using namespace std;

int N, M, L;
int P[100], T[100], V[100];
double prob[55][100];

void calc(int x)
{
	double p = P[x] / 100.0;
	double rest[55][55];

	for(int i=0; i<=M; i++)
	for(int j=0; j<=M; j++)
		rest[j][i] = 0.0;

	rest[0][0] = 1.0;

	for(int i=0; i<M; i++)
	for(int j=0; j<M; j++)
	{
		rest[i+1][j] += rest[i][j] * (1-p);
		rest[i+1][j+1] += rest[i][j] * p;
	}


	for(int i=0; i<=M; i++)
		prob[i][x] = rest[M][i];
}

void solve()
{
	for(int i=0; i<N; i++)
		calc(i);

	for(int i=0; i<N; i++)
	{
		double res = 0.0;
		for(int j=0; j<=M; j++)
		{
			double p = prob[j][i];
			double ltm = 1.0*L/V[i] + (T[i]*j);

			for(int k=0; k<N; k++)
			{
				if(k==i) continue;
				
				double t = 0.0;
				for(int l=0; l<=M; l++)
				{
					double rtm = 1.0*L/V[k] + (T[k]*l);
					if(ltm < rtm) t += prob[l][k];
				}

				p *= t;
			}

			res += p;
		}

		cout.setf(ios::fixed);
		cout.precision(11);
		cout << res << endl;
	}
}



int main()
{
	
	cin >> N >> M >> L;

	for(int i=0; i<N; i++)
		cin >> P[i] >> T[i] >> V[i];

	solve();

}