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<cstdio>
#include<cmath>
int n,m,l,p[100],t[100],v[100];
double time[100][51],poss[100][51],c[2][51],ans,sum[100][51];
int main(){
        scanf("%d%d%d",&n,&m,&l);
        c[0][0]=1;
        for (int i=1;i<=m;i++){
                c[i%2][0]=c[i%2][i]=1;
		for (int j=1;j<i;j++) c[i%2][j]=c[(i+1)%2][j-1]+c[(i+1)%2][j];
        }
        for (int i=0;i<n;i++){
                scanf("%d%d%d",&p[i],&t[i],&v[i]);
                time[i][0]=(double)l/v[i];
                for (int j=1;j<=m;j++) time[i][j]=time[i][j-1]+t[i];
                for (int j=0;j<=m;j++) poss[i][j]=c[m%2][j]*pow((double)p[i]/100,j)*pow(1-(double)p[i]/100,m-j);
		sum[i][m]=poss[i][m];
		for (int j=m-1;j>=0;j--) sum[i][j]=sum[i][j+1]+poss[i][j];
        }
	for (int i=0;i<n;i++){
		ans=0;
		for (int j=0;j<=m;j++){
			double cur=1,cur_t=time[i][j];
			for (int k=0;k<n;k++) if (i!=k){
				int l=0,r=m,mid,res;
				while (l<=r){
					mid=l+r>>1;
					if (time[k][mid]>cur_t){
						res=mid;
						r=mid-1;
					}	
					else l=mid+1; 
				}
				if (time[k][res]>cur_t) cur*=sum[k][res];
				else cur=0;
			}
			ans+=cur*poss[i][j];
		}
		printf("%f\n",ans);
	}
	return 0;
}