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:
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define EPS 1e-12
using namespace std;

int i;
double pr[111][55], tim[111][55];
double t[111], p[111], v[111];
int n, m;
long long comb[55][55];
double l;

double compute_t(int x, double v, double t) {
  return 1. * l / v + t * x;
}

double compute_p(int x, double p) {
  return (1.0 * comb[m][x]) * pow(p, x) * pow(1.0 - p, m - x);
}

void Pascal()
{
  for(int i = 0; i <= 50; i++) {
    comb[i][0] = 1;
    for(int j = 1; j <= i; j++) {
      comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
    }
  }
}

int main() {
  Pascal();         
  cin >> n >> m >> l;
  for(int i = 1; i <= n; i++) {
    cin >> p[i] >> t[i] >> v[i];
    p[i] /= 100.0;
  }

  for(int i = 1; i <= n; i++)
    for(int j = 0; j <= m; j++) {
      tim[i][j] = compute_t(j, v[i], t[i]);
      pr[i][j] = compute_p(j, p[i]);
    }
                              
  cout << fixed << setprecision(6);
  
  
  for(int i = 1; i <= n; i++) {
    double ans = 0.;
    for(int x = 0; x <= m; x++) {
      double prob = 1.;
      for(int j = 1; j <= n; j++) {
        if(j == i) continue;
        int ind = m;
        double pp = 0.;
        while(ind >= 0 && tim[j][ind] > tim[i][x]) pp += pr[j][ind--];
        prob *= pp;                                  
      }
     // cout << i << ' ' << x << ' ' << prob << '\n';
      ans += prob * pr[i][x];
    }
    cout << ans << '\n';
  }

  return 0;
}