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<cmath>
#include<cstdio>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const double EPS=1e-7;
const double INF=1e77;

const int N_BIN=50;
double nCr[N_BIN+1][N_BIN+1];
void binom(){
	rep(n,N_BIN+1) nCr[n][0]=1;
	rep(n,N_BIN) rep(r,n+1) nCr[n+1][r+1]=nCr[n][r+1]+nCr[n][r];
}

int main(){
	binom();

	int n,m,L; scanf("%d%d%d",&n,&m,&L);
	double p[100];
	int rest[100],v[100];
	rep(i,n) scanf("%lf%d%d",p+i,rest+i,v+i), p[i]/=100;

	double t[100][51],q[100][51];
	rep(i,n) rep(j,m+1) {
		if(v[i]>0) t[i][j]=(double)L/v[i]+j*rest[i];
		else       t[i][j]=INF;
		q[i][j]=nCr[m][j]*pow(p[i],j)*pow(1-p[i],m-j);
	}

	rep(i,n){
		if(v[i]==0){ puts("0"); continue; }

		double ans=0;
		rep(j,m+1){
			double pr=q[i][j];
			rep(k,n) if(i!=k) {
				double tmp=0;
				rep(l,m+1) if(t[k][l]<0 || t[i][j]+EPS<t[k][l]) tmp+=q[k][l];
				pr*=tmp;
			}
			ans+=pr;
		}
		printf("%.15f\n",ans);
	}

	return 0;
}