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>
#include<cassert>
#include<climits>
#include<cstdio>
#include<iomanip>

#define REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)
#define IINF (INT_MAX)
#define MAX 500

using namespace std;

double C[MAX][MAX];
 
void makeC()
{
  C[0][0] = 1;
  rep(i,MAX-1)rep(j,i+1)
    {
      C[i+1][j] += C[i][j];
      if(j+1<MAX)C[i+1][j+1] += C[i][j];
    }
}


int N,M;
double L;


int main(){

  makeC();

  cin >> N >> M >> L;
  vector<double> P(N),T(N),V(N),PB[N],TM[N];

  //cout << "C[" << M << "][" << 0 << "] = " << C[M][0] << endl;

  rep(i,N){
    cin >> P[i] >> T[i] >> V[i];
    double p = P[i] / 100.0;
    rep(j,M+1){
      TM[i].push_back( L / V[i] + j * T[i] );
      PB[i].push_back( C[M][j] * pow(p,j) * pow(1-p,M-j) );
    }
  }

  rep(sp,N){
    double ans = 0;
    rep(i,TM[sp].size()){
      double cost = 1;
      rep(gp,N){
	double cost2 = 0;
	if(sp == gp)continue;
	rep(j,TM[gp].size()){
	  if(TM[sp][i] >= TM[gp][j]){
	    cost2 += PB[gp][j];
	  } else break;
	}
	cost *= ( 1 - cost2 );
      }
      ans += cost * PB[sp][i];
    }
    printf("%.10f\n",ans);
  }

  return 0;
}