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 <queue>
#include <string>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <cstdlib>
#include <map>
#include <memory.h>
#include <vector>
using namespace std;

#define rep(i, n) for(int i = 0; i < n; i++)
const double EPS = 1e-12;
const int INF = 1 << 29;

double cprob[100][100];
double P[110];
double T[110];
double V[110];

double comb(int n, int r){
  double res = 1.0;
  r = min(n - r, r);
  rep(i, r){
    res = res * (n - i) / (i + 1);
  }
  return res;
}

double getp(int M, int j, int id){
  double res = comb(M, j);
  rep(i, j) res *= P[id] / 100.0;
  rep(i, M - j) res *= (1 - P[id] / 100.0);
  return res;
}

int main(){
  int N, M, L;
  double prob, p, p2;
  while(cin >> N >> M >> L && (N || M || L)){
    rep(i, N){
      cin >> P[i] >> T[i] >> V[i];
    }
    rep(i, N)rep(j, M + 1) cprob[i][j] = getp(M, j, i);
    rep(i, N){
      prob = 0.0;
      rep(j, M + 1){
	p = getp(M, j, i);
	double need = (double)L / V[i] + j * T[i];
	rep(k, N){
	  double p1 = 0.0;
	  if(i == k) continue;
	  rep(l, M + 1){
	    p2 = cprob[k][l];
	    double ene = (double)L / V[k] + l * T[k];
	    if(ene > need) {
	      p1 += p2;
	    }
	  }
	  p *= p1;
	}
	prob += p;
      }
      cout << fixed << setprecision(12) << prob << endl;
    }
    
  }
  return 0;
}