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>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

vector<vector<double>> c;

double pp(int m, int k, double p){
  double ans=c[m][k];
  rep(i,k) ans*=p;
  rep(i,m-k) ans*=1-p;
  return ans;
}

int main(){
  c=vector<vector<double>>(51,vector<double>(51));
  c[0][0]=1;
  repp(i,1,51){
    c[i][0]=1,c[i][i]=1;
    repp(j,1,i){
      c[i][j]=c[i-1][j-1]+c[i-1][j];
    }
  }
  int n,m;
  double l;
  cin >> n >> m >> l;
  vector<vector<double>> time(n,vector<double>(m+1)),prob(n,vector<double>(m+1));
  double p,t,v;
  rep(i,n){
    cin >> p >> t >> v;
    rep(j,m+1){
      time[i][j]=l/v+t*j;
      prob[i][j]=pp(m,j,p/100);
    }
  }
  cout << fixed << setprecision(10);
  rep(i,n){
    double ans=0;
    rep(j,m+1){
      double tmp=prob[i][j];
      rep(k,n){
        if(i==k) continue;
        double tmpp=0;
        per(l,m+1){
          if(time[k][l]<=time[i][j]) break;
          tmpp+=prob[k][l];
        }
        tmp*=tmpp;
      }
      ans+=tmp;
    }
    cout << ans << endl;
  }
  return 0;
}