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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

double combination(int a, int b){
  double ret = 1;
  for(int i = 0; i < b; ++i){
    ret *= (double)(a-i)/(i+1);
  }
  return ret;
}

int main(){
  int N, M, L;
  cin >> N >> M >> L;
  int P[N], T[N], V[N];
  for(int i = 0; i < N; ++i){
    cin >> P[i] >> T[i] >> V[i];
  }
  double prob[N][M+1];

  for(int i = 0; i < N; ++i) prob[i][0] = 1.0;
  for(int i = 0; i < N; ++i){

    for(int j = 1; j <= M; ++j){
      prob[i][j] = prob[i][j-1]*P[i]/100.0*(M-j+1)/j;
    }
    double t = 1.0;
    for(int j = M; j >= 0; --j){
      prob[i][j] *= t;
      t *= (1.0 - (double)P[i]/100.0);
    }
  }
  double acum[N][M+2];
  for(int i = 0; i < N; ++i) acum[i][M+1] = 0;
  for(int i = 0; i < N; ++i){
    for(int j = M; j >= 0; --j){
      acum[i][j] = acum[i][j+1] + prob[i][j];
    }
  }
  vector< vector<double> > time(N,vector<double>(M+1));
  for(int i = 0; i < N; ++i) time[i][0] = (double)L/V[i];
  for(int i = 0; i < N; ++i){
    for(int j = 1; j <= M; ++j){
      time[i][j] = time[i][j-1] + T[i];
    }
  }
  for(int i = 0; i < N; ++i){
    double ans = 0.0;
    for(int j = M; j >= 0; --j){
      double t = prob[i][j];
      for(int k = 0; k < N; ++k){
        if(k == i) continue;
        int id = upper_bound(time[k].begin(),time[k].end(),time[i][j]) - time[k].begin();
        if(id > M){
          t = 0;
          break;
        }
        t *= acum[k][id];
      }
      ans += t;
    }
    printf("%.12f\n",ans);
  }
  return 0;
}