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<cmath>
#include<cstdio>
using namespace std;
#define REP(i,b,n) for(int i=b;i<n;i++)
#define rep(i,n)   REP(i,0,n)
const int N = 101;
const int M = 51;
double dp[N][M][M];//probability: who,ith place,num of QK
double t[N][M];//num: who,num of QK,

void solve(int n,int I,double prob){
  dp[I][0][0]=1-prob;
  dp[I][0][1]=prob;
  REP(i,1,n){
    REP(j,0,n+1){
      dp[I][i][j]=0;
      dp[I][i][j]+=dp[I][i-1][j]*(1-prob);
      if (j-1 >= 0){
        dp[I][i][j]+=dp[I][i-1][j-1]*prob;
      }
    }
  }
}

int main(){
  int n,m,l;
  while(cin>>n>>m>>l){
    double p[n],ti[n],v[n];
    rep(i,n){
      cin>>p[i]>>ti[i]>>v[i];
      p[i]=p[i]/100.0;
      double tttt=l/v[i];
      rep(j,m+1){
        t[i][j]=tttt+j*ti[i];
      }
    }
    rep(i,n)solve(m,i,p[i]);
    /*
    rep(i,n){
      rep(j,m+1){
        cout << t[i][j] <<"," << dp[i][m-1][j] <<" ";
      }
      cout << endl;
    }
    */
    if (m == 0){
      rep(i,n){
        double a=1;
        rep(j,n){
          if (i == j)continue;
          if (v[i] <= v[j])a=0;
        }
        printf("%.12lf\n",a);
      }
    }else {
      rep(i,n){//
        double sum=0;
        rep(k,m+1){
          double prob=1;
          rep(j,n){
            if (i == j)continue;
            double ahoge=0;
            rep(l,m+1){
              if (t[i][k] < t[j][l])ahoge+=dp[j][m-1][l];
            }
            prob*=ahoge;
          }
          sum+=prob*dp[i][m-1][k];
        }
        printf("%.12lf\n",sum);
      }
      //cout << endl;
#ifdef DEBUG
      printf("end\n");
#endif
    }

  }
  return false;
}