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)

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

void solve() {
	for_(i,0,N) {
		double dp[55][55] = { 0.0 }, pi = (double)P[i] / 100.;
		dp[0][0] = 1.0;
		
		for_(j,0,M) for_(k,0,M) {
			dp[j+1][k] += dp[j][k] * (1.0 - pi);
			dp[j+1][k+1] += dp[j][k] * pi;
		}
		
		for_(k,0,M+1) p[i][k] = dp[M][k];
	}
	
	for_(i,0,N) {
		double ans = 0.0;
		
		if (V[i] > 0) {
			for_(j,0,M+1) {
				double t1 = (double)L / (double)V[i] + (double)(j * T[i]);
				double mul = 1.0;
				
				for_(ii,0,N) {
					if (i != ii && V[ii] > 0) {
						double mii = 0.0;								
						double t2 = (double)L / (double)V[ii] + (double)(M * T[ii]);
						
						for_rev(k,M,0) {
							if (t2 - t1 < 1e-9) break;
							mii += p[ii][k];
							t2 -= (double)T[ii];
						}
					
						mul *= mii;
					}
				}
				
				ans += mul * p[i][j];
			}
		}
		
		printf("%.9lf\n", (N == 1) ? 1.0 : ans);
	}
}

int main() {
	cin >> N >> M >> L;
	for_(i,0,N) cin >> P[i] >> T[i] >> V[i];
	
	solve();
	
	return 0;
}