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>
using namespace std;

double P[100], T[100], V[100];
double ncr[55][55];
int main(){
	ncr[0][0] = 1.0;
	for(int i = 0 ; i < 50 ; i++)
		for(int j = 0 ; j < 50 ; j++)
			ncr[i+1][j] += ncr[i][j],
			ncr[i+1][j+1] += ncr[i][j];
	
	int N,M,L;
	cin >> N >> M >> L;
	for(int i = 0 ; i < N ; i++){
		cin >> P[i] >> T[i] >> V[i];
		P[i] /= 100.0;
		//P[i] = 1 - P[i];
	}
	for(int i = 0 ; i < N ; i++){
		double sum2 = 0;
		for(int l = 0 ; l <= M ; l++){
			double myTime = L / V[i] + l * T[i];
			double prob1 = ncr[M][l] * pow(P[i],l) * pow(1-P[i],M-l);
			//cout << i << " " << l << " " << prob1 << endl;
			double OKProb = 1.0;
			for(int j = 0 ; j < N ; j++){
				if( i == j ) continue;
				double sum = 0;
				for(int k = 0 ; k <= M ; k++){
					double time = L / V[j] + k * T[j];
					double prob2 = ncr[M][k] * pow(P[j],k) * pow(1-P[j],M-k);
					if( time > myTime + 1e-10 ){
						sum += prob2;
					}
				}
				OKProb *= sum;
			}
			sum2 += OKProb * prob1;
		}
		printf("%.10lf\n",sum2);
	}
}