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 <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

struct Runner{
  double p, t, v;
  Runner(double p, double t, double v) : p(p), t(t), v(v){}  
};

int main(){
  int N, M, L;
  cin >> N >> M >> L;
  
  vector<Runner> r;
  
  for(int i = 0 ; i < N ; i++){
    double p, t, v;
    cin >> p >> t >> v;
    r.push_back(Runner(p/100.0, t, v));
  }
  
  double dp[55][110][55]; // iテァツ閉ェテァツ崢ョテ」ツ?ァ j テ」ツ??kテ・ツ崢?テ、ツシツ妥」ツつ?ァツ「ツコテァツ篠?
  memset(dp, 0, sizeof(dp));
  for(int i = 0 ; i < N ; i++) dp[0][i][0] = 1;
  
  for(int i = 1 ; i <= M ; i++){
    for(int j = 0 ; j < N ; j++){
      for(int k = 0 ; k <= M ; k++){
	dp[i][j][k] = (r[j].p * dp[i-1][j][k-1]) + ((1.0-r[j].p) * dp[i-1][j][k]);
      }
    }
  }
  
  double sum[110][55];
  for(int i = 0 ; i < N ; i++){
    sum[i][0] = (double)(L / r[i].v);
    for(int j = 1 ; j <= M ; j++){
      sum[i][j] = sum[i][j-1] + r[i].t;
    }
  }
  
  double sum2[110][55];
  for(int i = 0 ; i < N ; i++){
    sum2[i][0] = dp[M][i][0];
    for(int j = 1 ; j <= M ; j++){
      sum2[i][j] = sum2[i][j-1] + dp[M][i][j];
    }
  }
  
  for(int i = 0 ; i < N ; i++){
    double ans = 0.0;

    for(int j = 0 ; j <= M ; j++){
      double p = dp[M][i][j];
      for(int ii = 0 ; ii < N ; ii++){
	if(i == ii) continue;

	int index = upper_bound(sum[ii], sum[ii]+M+1, sum[i][j]) - sum[ii];

	double s = sum2[ii][M] - sum2[ii][index-1];
	p *= s;	
      }
      ans += p;
    }
    printf("%.8f\n", ans);
  }
  return 0;
}