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 <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
#define rep1(i,n) for(int i=1;i<=(n);++i)
#define all(c) (c).begin(),(c).end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl;
double eps=1e-9,inf=1e10;
typedef pair<double,double> P;
typedef pair<P,int> PP;
vector<PP> vc;
int main(){
	int n,m,l;
	cin>>n>>m>>l;
	rep(i,n){
		double p,t,v;
		cin>>p>>t>>v;
		p/=100;
		if(p==1){
			double tt;
			if(v==0) tt=inf;
			else tt=l/v+m*t;
			vc.pb(PP(P(tt,1),i));
			//cout << "i" << i << " " << tt << " " << 1 << endl;
			continue;
		}
		if(p==0){
			double tt;
			if(v==0) tt=inf;
			else tt=l/v;
			vc.pb(PP(P(tt,1),i));
			//cout << "i" << i << " " << tt << " " << 1 << endl;
			continue;
		}
		double pro=1;
		rep(j,m) pro*=(1-p);
		rep(j,m+1){
			double tt;
			if(v==0) tt=inf;
			else tt=l/v+j*t;
			vc.pb(PP(P(tt,pro),i));
			//cout << "i" << i << " " << tt << " " << pro << endl;
			pro*=p/(1-p);
			pro*=(double)(m-j)/(j+1);
		}
	}
	sort(all(vc));
	double ans[100]={};
	rep(i,vc.size()){
		if(vc[i].fs.fs>1e9) break;
		double cnt[100]={};
		for(int j=i+1;j<vc.size();j++){
			if(vc[j].fs.fs<vc[i].fs.fs+eps) continue;
			cnt[vc[j].sc]+=vc[j].fs.sc;
		}
		double ret=1;
		rep(j,n){
			if(j==vc[i].sc) ret*=vc[i].fs.sc;
			else ret*=cnt[j];
		}
		ans[vc[i].sc]+=ret;
	}
	rep(i,n) printf("%.9f\n",ans[i]);
}