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;

#define for_(i,a,b) for(int i=(a);i<(b);++i)
#define for_rev(i,a,b) for(int i=(a);i>=(b);--i)
#define allof(a) (a).begin(),(a).end()
#define size_of(a) (int)(a).size()
#define minit(a,b) memset(a,b,sizeof(a))

int N, M, L, P[110], T[110], V[110];
double RP[110][55], RT[110][55];

int main() {
	cin >> N >> M >> L;
	for_(i,0,N) cin >> P[i] >> T[i] >> V[i];
	
	minit(RP, 0);
		
	for_(i,0,N) {
		RT[i][0] = V[i] > 0 ? (double)L / (double)V[i] : 1e9;
		for_(j,0,M) RT[i][j + 1] = RT[i][j] + (double)T[i];
	}
	
	for_(i,0,N) {
		vector< double > p(M+1, 1.0), pm(M+1, 1.0);
		
		for_(k,0,M) {
			p[k + 1] = p[k] * (double)P[i] / 100.0;
			pm[k + 1] = pm[k] * (1.0 - (double)P[i] / 100.0);
		}
		
		long long mCk = 1;
		
		for_(k,0,M + 1) {
			RP[i][k] =  (double)mCk * p[k] * pm[M - k];
			mCk *= (M - k);
			mCk /= k + 1;
		}
		
		for_rev(k,M,1) RP[i][k-1] += RP[i][k];
	}
		
	for_(i,0,N) {
		if (V[i] == 0) {
			cout << "0.000000000" << endl;
			continue;
		}
		
		double ans = 0.0;
		
		for_(k,0,M+1) {
			double t = RT[i][k];
			double p = RP[i][k] - RP[i][k + 1];
						
			for_(ii,0,N) {
				if (ii == i || V[ii] == 0) continue;
				int id = upper_bound(RT[ii], RT[ii] + M + 1, t) - RT[ii];
				p *= RP[ii][id];
			}
			
			ans += p;
		}
		
		cout << setprecision(9) << setiosflags(ios::fixed) << ans << endl;
	}
}