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 <bits/stdc++.h>

#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;

typedef long double ld;

int main() {
  int n,m;
  double l;
  cin>>n>>m>>l;
  vector<vector<ld>> prob(n, vector<ld>(m+1));
  vector<vector<ld>> tm(n, vector<ld>(m+1));
  REP(i,n) {
    ld p,t,v;
    cin>>p>>t>>v;
    p /= 100.0;
    prob[i][0] = 1.0;
    REP(j,m) {
      vector<ld> tmp(m+1);
      tmp[0] = prob[i][0] * (1-p);
      REP(k,m)
        tmp[k+1] = prob[i][k] * p + prob[i][k+1] * (1-p);
      swap(prob[i], tmp);
    }
    REP(j,m+1)
      tm[i][j] = l/v + t*j;
  }
  REP(i,n) {
    ld psum = 0.0;
    REP(j,m+1) {
      ld T = tm[i][j];
      ld p = prob[i][j];
      REP(k,n) {
        if(i==k) continue;
        ld sp = 0.0;
        REP(l,m+1){
          if (tm[k][l] > T) sp += prob[k][l];
        }
        p *= sp;
      }
      psum += p;
    }
    cout << setprecision(10) << fixed << psum << endl;
  }
  return 0;
}