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 <vector>
#include <algorithm>
#include <cmath>
using namespace std;

#define EPS (1e-10)
#define EQ(a,b) (abs((a)-(b))<EPS)

struct runner{
	int t,v;
	double p;
};

const int INF=1000000000;

vector<runner> vr;
double mem[51][51][101];
// kÍxe·éêÌ
double calcPer(int n,int k,int idx){
	if(!EQ(mem[n][k][idx],INF))
		return mem[n][k][idx];
	double p = vr[idx].p;
	if(k==0)
		return 1.0;
    return mem[n][k][idx]=(((double)n/k)*calcPer(n-1,k-1,idx)*p);
}
int main(){
	int n,m,L;
	for(int i = 0; i < 51; i++)
		for(int j = 0; j < 51; j++)
			for(int k = 0; k < 101; k++)
				mem[i][j][k]=INF;
	cin>>n>>m>>L;
	for(int i = 0; i < n; i++){
		runner r;
		int p;
		cin>>p>>r.t>>r.v;
		r.p=(double)p/100;
		vr.push_back(r);
	}
	// êl¸ÂßÄ¢­
	for(int i = 0; i < n; i++){
		double sumPer=0.0;
		if(vr[i].v!=0){
			// ©ªª¢­ÂÌêÅxe·é©
			for(int j = 0; j <= m; j++){
				// »ÝÌxeÅxe·ém¦
				double cper=calcPer(m,j,i)*pow((double)(1-vr[i].p),m-j);
                double selfTime=((double)L/vr[i].v)+vr[i].t*j;
				// »ê¼êÌGÉÂ¢Ä²×é
				for(int k = 0; k < n; k++){
					// ©ª©gÜ½ÍA¬xª0Ìi[Í³·é
					if(i==k||EQ(vr[k].v,0.0))
						continue;
					double enemPer=1.0;
					// èª¢­ÂÈãÌêÅxeµ½çAÔÉ¤©ð`FbN
					for(int l = 0; l <= m; l++){
                        // GÌBðvZ
                        double enemyTime=((double)L/vr[k].v)+vr[k].t*l;
						// àµGÌBÔª©ªÌBÔðãñÁ½çI¹
						if(!EQ(enemyTime,selfTime)&&enemyTime>selfTime)
                            break;
                        // ±ÌÌêÅxe·ém¦
                        double p=calcPer(m,l,k)*pow((double)(1-vr[k].p),m-l);
                        // ¡ñÌxeðæéêÍÂ±ÆªoÈ¢ÌÅA}CiX
						enemPer-=p;
					}
					// »ÝcÁÄ¢ém¦ªAGª©ªæèx­B·ém¦
					cper*=enemPer;
				}
				// cperª»Ìm¦
				sumPer+=cper;
			}
		}
		printf("%.7f\n",sumPer);
	}
	return 0;
}