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 <stdio.h>
#include <algorithm>
#include <utility>
#define inf 10000000000

using namespace std;
typedef pair<double, double> P;

int N, M, L;
double p[105], T[105], V[105];
P table[105][55];

double pow(double a, int n)
{
	if(n == 0) return 1.0;
	if(n % 2) return a * pow(a, n-1);
	else{
		double t = pow(a, n/2);
		return t * t;
	}
}

double comb(double n, double k)
{
	double ret = 1.0;
	for(int i = 0; i < k; i++){
		ret *= (n-i)/(k-i);
	}
	return ret;
}

int main(void)
{
	cin >> N >> M >> L;
	
	int pp;
	for(int i = 0; i < N; i++){
		cin >> pp >> T[i] >> V[i];
		p[i] = pp / 100.0;
	}
	
	for(int i = 0; i < N; i++){
		for(int j = 0; j <= M; j++){
			table[i][j] = make_pair( L/V[i] + j*T[i], comb(M, j) * pow(p[i],j) * pow(1.0-p[i],M-j) );
			if(j != 0) table[i][j].second += table[i][j-1].second;
		}
	}
	
	P *ix;
	double ans, prob;
	for(int i = 0; i < N; i++){
		ans = 0.0;
		for(int j = 0; j <= M; j++){
			
			prob = table[i][j].second;
			if(j != 0) prob -= table[i][j-1].second;
			
			for(int k = 0; k < N; k++){
				if(i == k) continue;
				ix = lower_bound(table[k], table[k] + M+1, make_pair(table[i][j].first, 1.1));
				if(ix != table[k]){
					prob *= 1.0 - ix[-1].second;
				}
			}
			ans += prob;
		}
		printf("%.8f\n", ans);
	}
	
	return 0;
}