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 fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
typedef long long LL;
typedef pair<double,int> P;
const LL mod=1000000007;
const LL LINF=1LL<<62;
const int INF=1<<30;
int dx[]={0,0,1,1,1,-1,-1,-1};
int dy[]={1,-1,1,0,-1,1,0,-1};

vector<double> fact;
 
double combi(int n,int r){
    if(n<r||n<0||r<0) return 0;
    return fact[n]/fact[n-r]/fact[r];
}
 
 
void set_combi(){
    double s=1.0;
    fact.push_back(1.0);
    for(int i=1;i<=50;i++){
        s*=i;
        fact.push_back(s);
    }
}

double fpow(double a, int n){
    double x = 1;
    while(n > 0){
        if(n&1){
            x=x*a;
        }
        a=a*a;
        n >>= 1;
    }
    return x;
}

int main(){
    set_combi();
    int n,m,l;cin >> n >> m >> l;
    vector<LL> p(n),t(n),v(n);
    for (int i = 0; i < n; i++) {
        cin >> p[i] >> t[i] >> v[i];
    }
    vector<vector<double>> d(n,vector<double>(m+1,0.0));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= m; j++) {
            d[i][j] = combi(m,j) * fpow(p[i]/100.,j) * fpow((100-p[i])/100.,m-j);
        }
    }
    for (int i = 0; i < n; i++) {
        double ans = 0.0;
        for (int j = 0; j <= m; j++) {
            double pp = d[i][j];
            for (int k = 0; k < n; k++) {
                if(i==k) continue;
                double tmp = 0.0;
                for (int c = 0; c <= m; c++) {
                    if(c * t[k] * v[k] * v[i] + l * v[i] <= j * t[i] * v[k] * v[i] + l * v[k]) continue;
                    tmp += d[k][c];
                }
                pp *= tmp;
            }
            ans += pp;
        }
        cout << setprecision(30) << fixed << ans << endl;
    }
    return 0;
}