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>
#define rep(i,n) for(int i = 0; i < n; i++)
#define INF 100000000
#define EPS 1e-10
#define MOD 1000000007
using namespace std;

int n, m;
double l;
double p[100], t[100], v[100];
double cb[100][51];

void init(){
	rep(u,n){
		rep(i,m+1){
			int j = m-i;
			cb[u][i] = 1.0;
			int cnt1 = i, cnt2 = j;
			for(int k = 1; k <= i; k++){
				cb[u][i] = cb[u][i]*(m-k+1)/k;
				while(cb[u][i] > 1.0){
					if(cnt1 > 0){
						cb[u][i] *= p[u];
						cnt1--;
					} else if(cnt2 > 0){
						cb[u][i] *= (1-p[u]);
						cnt2--;
					}
					if(cnt1 == 0 && cnt2 == 0) break;
				}
			}
			rep(k,cnt1) cb[u][i] *= p[u];
			rep(k,cnt2) cb[u][i] *= (1-p[u]);
		}
	}
}

int main(){
	cin >> n >> m >> l;
	rep(i,n){
		cin >> p[i] >> t[i] >> v[i];
		p[i] *= 0.01;
	}
	init();
	rep(i,n){
		double pos;
		//double ti;
		double ans = 0.0;
		if(v[i] == 0.0){
			puts("0.00000000000");
			continue;
		}
		for(int j = 0; j <= m; j++){
			pos = cb[i][j];
			rep(k,n){
				if(k == i) continue;
				if(v[k] == 0.0) continue;
				if(t[k] == 0.0){
					if((int)l*(int)v[k]-(int)l*(int)v[i]+(int)t[i]*j*(int)v[i]*(int)v[k] >= 0) pos *= 0.0;
					else pos *= 1.0;
					continue;
				}
				int c;
				if((int)l*(int)v[k]-(int)l*(int)v[i]+(int)t[i]*j*(int)v[i]*(int)v[k] < 0) c = 0;
				else c = ((int)l*(int)v[k]-(int)l*(int)v[i]+(int)t[i]*j*(int)v[i]*(int)v[k])/((int)t[k]*(int)v[i]*(int)v[k]) + 1;
				double tmp = 0.0;
				for(int l = c; l <= m; l++){
					tmp += cb[k][l];
				}
				pos *= tmp;
			}
			ans += pos;
		}
		printf("%.9f\n", ans);
	}
}