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>
using namespace std;

int N,M,L;
int P[111],T[111],V[111];

double Time[111][111];
double pa[111][111];
double ncr[111][111];

const double EPS = 1E-9;
const double INF = 1E50;

void getncr(){
  for(int i=0;i<=100;i++) ncr[i][i] = ncr[i][0] = 1LL;
  for(int i=2;i<=100;i++)
    for(int j=1;j<i;j++)
      ncr[i][j] = ncr[i-1][j] + ncr[i-1][j-1];
      	
}

int main(){
  cin >> N >> M >> L;
  for(int i=0;i<N;i++){
    cin >> P[i] >> T[i] >> V[i];
  }

  getncr();

  for(int i=0;i<N;i++){
    double ka = 1.0f;

    if( P[i] == 100 ){
      for(int j=0;j<=M;j++){
	if( V[i] == 0 )
	  Time[i][j] = INF;
	else
	  Time[i][j] = (double)L / (double)V[i] + (double)T[i]*(double)j;	
	pa[i][j] = 0.0f;
      }      
      pa[i][M] = 1.0f;
    } else {
      for(int l=0;l<M;l++){
	ka *= (double)(100-P[i])/100.0;
      }
      //  cout <<i << endl;
      for(int j=0;j<=M;j++){
	if( V[i] == 0 )
	  Time[i][j] = INF;
	else
	  Time[i][j] = (double)L / (double)V[i] + (double)T[i]*(double)j;
	
	if( ka > 1.0 - EPS ) pa[i][j] = 1.0f;
	else pa[i][j] = ka*ncr[M][j];  
	
	ka /= (double)(100-P[i])/100.0;
	ka *= (double)P[i]/100.0;       
      }
    }
  }

  for(int i=0;i<N;i++){
    double res = 0.0f;
    for(int j=0;j<=M;j++){

      double H = Time[i][j];
      double ret = 1.0f;
      for(int k=0;k<N;k++){
	if( i == k ) continue;
	double kk = 0.0f;

	for(int l=0;l<=M;l++){
	  // cout << H << " " << Time[k][l] << " " << pa[k][l] << endl;
	  if( Time[k][l] > H+EPS ){
	    //      cout << pa[k][l] << endl;
	    kk += pa[k][l];
	  }
	}

	//	cout << "k= " << k << " "<< kk << endl;
	ret *= kk;
      }
      // cout << "i , j = " << i << " "<< j << " "<< ret << " * " << pa[i][j] << " "<< ret*pa[i][j] << endl;
      res += ret*pa[i][j];
    }

     printf("%.10lf\n",res);
  }

}